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Understanding rates of change

Formula for rate of change, function rate of change, graphical rates of change, real-world examples of rate changes, key takeaways on rate of change, rates of change: key definitions, formula & examples.

Are you aware that the word 'change' is among the most influential in political campaigns?

Upon contracting Covid-19, it's possible to calculate how quickly the virus spreads over a certain time frame.

Through this article, the concept of rate of change and its practical uses will be made clear.

The rate of change refers to the ratio that exists between changes in two different quantities.

Often referred to as the slope or gradient, this occurs when comparing alterations across two quantities.

Rate of change is a fundamental concept used in creating various formulas, including those for velocity and acceleration, indicating the level of activity amidst variations in contributing quantities.

Consider a vehicle traveling A meters in n seconds.

Moving from point A, it traverses an additional B distance by the mth second, showing variances between distances A to B and time from nth to mth second.

Dividing these variances yields the rate of change.

Defining change in mathematics

In the realm of mathematics, change occurs with any increase or decrease in a quantity's value.

This suggests that the change might be positive, negative, or even zero if the quantity remains unchanged.

If you currently possess 5 oranges and later find you have 8, has there been a change? Indeed, as the count of oranges has risen by 3, indicating a positive change.

Conversely, if you start with 5 oranges and end up with just one, this implies a loss of 4 oranges, representing a negative change.

It's important to realize that change is essentially the differential in quantities, calculated as,

Δ Q = Q f - Q i

∆ Q indicates the change in quantity,

Q i signifies the initial quantity,

Q f represents the quantity's final value.

A positive ΔQ indicates a positive change, whereas a negative ΔQ suggests a decrease.

Now that the concept of change is clear, let's proceed to evaluate the rate of change.

To determine the rate of change, we divide the alterations in quantities, implying,

rate of change = change in quantity change in quantity

Building on this formula, we'll use graph directions for guidance. Assume modifications happen along the x-axis (horizontal) and y-axis (vertical).

In the x-direction, we describe a change as

Δ x = x f - x i

∆ x denotes the variance in the x-direction,

x i indicates the starting point,

x f marks the endpoint on the x-axis.

For changes in the y-direction, we note as,

Δ y = y f - y i

∆ y characterizes the shift in the y-direction,

y i is the initial vertical position,

y f is the final point on the y-axis.

So, we articulate the rate of change formula as,

r a t e o f c h a n g e = Δ y Δ x = y f - y i x f - x i r a t e o f c h a n g e = y f - y i x f - x i

If initially a value was observed at 5 units along x and 3 units along y, and later changed to 8 units x-wise and 4 units y-wise, what's the rate of change?

Given the data, we find

x i as 5, x f as 8

y i as 3, y f as 4

rate of change = y f - y i x f - x i = 1 3

The function's rate of change measures how the outcome of a function varies with the change in the input quantity.

Let's assign u as a function of w, depicted as

w = f ( u ) .

This rate demonstrates how w shifts in response to changes in u, provided w depends on u.

The variation in u is defined as

Δ u = u f - u i

∆ u represents the alteration in u ,

u i marks the original value of u ,

u f is the resulting value of u ,

Furthermore, the modification in w is articulated by

Δ w = w 1 - w 0

w = f ( u )

leading us to,

f ( Δ u ) = f ( u 1 - u 0 ) = f ( u 1 ) - f ( u 0 )

Hence, the formula for a function's rate of change is,

Δ w Δ u = f ( Δ u ) Δ u = f ( u f - u i ) u f - u i

The formula to assess a function's change rate is:

Δ y Δ x = f ( x f ) - f ( x i ) x f - x i

∆ x is the shift along the x-axis,

x i is the x-axis starting point,

x f is the endpoint on the x-axis,

∆ y is the vertical shift,

f ( x i ) is the initial x-position's function,

f ( x f ) is the concluding x-position's function.

Displaying rates of change graphically involves charting quantities on a graph, with three scenarios: zero, positive, and negative rate changes.

Instances of zero rate changes

A zero rate change is seen when there's no effect on the second quantity despite a change to the first. This occurs when

y f - y i = 0 .

The graphical representation below showcases the zero rate of change.

Observation of the horizontal orientation suggests x-values shift but y-values remain fixed, rendering the gradient as negligible.

Scenarios with positive rate changes

A positive rate change manifests when the quotient of changes yields a positive outcome. The slope's angle depends on the relative magnitude of change between the pair of quantities.

This infers a lenient slope if y-value alterations surpass those of x-values; conversely, a sharper gradient results when x-value changes exceed those of y-values.

The upward slope detected confirms the positive nature of the rate change. Inspect the images below to comprehend the concept further.

Instances of negative rate changes

Negative rate changes are identified when the quotient of variations results in a negative figure. This scenario emerges when one modification is negative and the other positive. Note, if both alterations yield negative figures, the rate change is considered positive, not negative.

The slope’s steepness again reflects the relative magnitude of change between the quantities, indicating a milder incline if y-value changes outweigh those of x-values, and a sharper descent if x-value changes overshadow those of y-values.

The downward direction of the slope solidifies the rate of change as negative. Examine these illustrations for a better grasp.

To compute the rate of change between two points (1,2) and (5,1):

a. Ascertain the variety of rate change.

b. Determine if the slope is mild or steep.

With x i = 1 , y i = 2 , x f = 5 , y f = 1 ,

To graphically plot, coordinates are marked within the plane.

Applying the calculation formula for rate change,

r a t e o f c h a n g e

a. Given our rate of change as -4, it categorizes as a negative rate change.

b. Observing a significant change towards the y-direction (4 positive steps) against a minor x-direction change (1 negative step), the graphed slope emerges as mild, as represented.

Various real-life applications involve rate of change, notably in calculating velocity. This example illuminates the concept:

To determine a vehicle’s average velocity, apply:

Average Velocity = Total Traversed Distance / Total Duration

With the vehicle covering 300m in the first 30 seconds and 500m in the subsequent 70 seconds, the total distance comes to 300m + 500m = 800m.

The summed time spans 30 seconds + 70 seconds = 100 seconds.

Thus, the car's average velocity computes as:

Average Velocity = 800m / 100s = 8 m/s

• Rate of change is established as the ratio between modifications among two quantities.
• Change is observed when a specific quantity either increases or decreases in value.
• The mechanism to figure the rate of change adopts the formula; Rate of change = (yf - yi) / (xf - xi)
• The rate of change concerning a function reflects how a function's output varies as its input does.
• To visually represent rates of change, one maps quantities onto a graph.

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Interpreting Rates of Change

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A rate of change is the ratio between the change in one quantity to the change in another quantity. Linear relationships have a constant rate of change.

Interpreting Rates of Change from Situations and Tables

Interpreting rates of change from graphs, interpreting rates of change from equations.

The tile pattern below is growing by three tiles per figure. Therefore, the tile pattern has a growth rate of 3.

A runner travels \(\SI{80}{feet}\) in \(\SI{8}{seconds}\). What is the runner's rate of change? Answer The runner is traveling \(\dfrac{\SI{80}{feet}}{\SI{8}{seconds}}\) or \(\SI{10}{feet.per.second}\).

Is the relationship between the inputs and outputs below linear? Input Output 0 11 3 5 5 1 8 -5 Answer If the rate of change between pairs of inputs and outputs in constant, then the relationship is linear. The rate of change between the pairs \((0,11)\) and \((3,5)\) is \(\dfrac{11-5}{0-} = \dfrac{6}{-3}=-2.\) The rate of change between the pairs \((3,5)\) and \((5,1)\) is \(\dfrac{5-1}{3-5} = \dfrac{4}{-2}=-2.\) The rate of change between the pairs \((5,1)\) and \((8,-5)\) is \(\dfrac{1-(-5)}{5-8} = \dfrac{6}{-3}=-2.\) The relationship is linear with a rate of change of -2.

Plants A and B are both growing at a constant rate. Which plant is growing faster?

The slope of a line shows the rate of change in a linear relationship.

For example, the graph below shows a rate of change of 10 liters per second. The slope of the line is \(\frac{10}{1}=10.\)

Which car has the greatest gas mileage? Answer Line A is the steepest so it has the greatest rate of change so car A has the greatest gas mileage. We can see that car A travels about 50 miles on one gallon of gas, car B travels about 30 miles on one gallon of gas, and car C travels about 10 miles on one gallon of gas.

A cylindrical container is being filled with water. The volume of water over time is shown below. What is the rate of change in the water volume?

Equations of lines in the form \(y=mx+b\) represent linear functions with constant rates of change. The rate of change in the relationship is represented by \(m.\)

The equation \(y=5,000x+12,0000\) represents the total number of miles on Zen's car, \(y,\) each year that she owned it, \(x.\) How many miles does Zen drive per year? Answer The growth rate in this equation is 5,000. Therefore, Zen drives 5000 miles per year. Zen's car had 12,000 miles on it when she purchased it.

In pattern A, the number of dots, \(y,\) in figure \(x\) is given by \(y=6x+7.\) In pattern B the number of dots, \(y,\) in figure \(x\) is given by \(y=7x+6.\) Which dot pattern is growing more quickly? Answer In pattern A, the growth rate is 6. In pattern B, the growth rate is 7. Therefore, pattern B is growing more quickly.

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Rate of Change Formula

The rate of change function is defined as the rate at which one quantity is changing with respect to another quantity. In simple terms, in the rate of change, the amount of change in one item is divided by the corresponding amount of change in another. Let us learn about the rate of change formula with a few examples in the end.

What Is the Rate of Change Formula?

The rate of change formula gives the relationship describing how one quantity changes in relation to the change in another quantity. The rate of change from the coordinates of y to coordinates of x can found out as Δy/ Δx = (y 2 - y 1 )/ (x 2 - x 1 ). For a linear function, the rate of change m is represented in the slope-intercept form for a line: y=mx+b whereas the rate of change of functions is otherwise defined as, (f(b)-f(a))/ b-a

• Formula 1: The basic formula for the rate of change is:

Rate of change = (Change in quantity 1) / (Change in quantity 2)

• Formula 2: Formulas of rate of change in algebra

Δy/ Δx = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

• Formula 3: Rate of change of functions

(f(b)-f(a))/ b-a

Applications of Rate of Change Formula

The rate of change tells us how something changes over time.

• Distance traveled by car in a certain amount of time.
• The current through an electrical circuit increases by some amperes for every volt of increased voltage.
• It is considered an important financial concept as well. It allows investors to spot security momentum and other trends.
• Work done per unit time.
• Work done and the number of people required for doing it

Let us have a look at a few solved examples to understand the rate of change formula better.

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Examples Using Rate of Change Formula

Example 1: Using the rate of change formula, calculate the rate of change for the following information in the table:

To find: Rate of change

Using the rate of change formula,

Rate of change = (Change in distance) / (Change in time)

Rate of change = (180-40) / (4-2)

Rate of change = (140) / (2)

Rate of change = 70

Answer: The rate of change is 70 or the rate of change of distance with time is 70 miles per hour.

Example 2: Calculate the rate of change for the following information in the table:

To find: Rate of change.

Using the rate of change Formula,

Rate of change = (Change in height of the tree) / (Change in days)

Rate of change = (7-4) / (140-50)

Rate of change = (3) / (90)

Rate of change = 1/30 = 0.033..

Answer: The rate of change is 0.033 or the rate of change of height of the tree with time in days is 0.033 inches per day.

Example 3: Find the rate of change for the situation: Ron completed 3 math assignments in one hour and Duke completed 6 assignments in two hours.

Rate of change = (Change in assignments done) / (Change in hours)

Rate of change = (6-3) / (2-1)

Rate of change = (3) / (1)

Rate of change = 3/1 = 3 assignments/hour

Answer: The rate of change is 3.0 or the rate of change of assignments done with time in hours is 3 assignments per hour.

FAQs on Rate of Change Formula

What is the formula for rate of change in math.

A rate of change formula is used to calculate the rate which describes how one quantity changes in relation to the change in another quantity. Thus, the formula for the rate of change is, ROC = (Change in quantity 1) / (Change in quantity 2)

What Is the Average Rate of Change Formula?

The average rate is the total change divided by the time taken for that change to occur. The way it is calculated is similar to how the average velocity of an object is calculated. For example, the average rate of change in a population of an area can be calculated with only the times and populations at the start and end of the period.

How To Use the Rate of Change Formula for Graphs?

The rate of change can be depicted and calculated using the formula for rate of change, that is \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\), commonly known as slope formula.

What Is the Instant Rate of Change Formula?

The instantaneous rate of change is defined as the change in the rate at a particular instant. It can be considered the same as the change in the derivative value at a specific point. For a graph, the instantaneous rate of change at a specific point is the same as the tangent line slope.

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AP®︎/College Calculus AB

Course: ap®︎/college calculus ab   >   unit 4.

• Interpreting the meaning of the derivative in context

Analyzing problems involving rates of change in applied contexts

At t = 5 ‍   seconds, the tank is being filled at a rate of 1 ‍   liter per second.

Lindsay is walking home from school. Her distance from school, in meters, after t ‍   minutes is modeled by the differentiable function D ‍   .

• (Choice A)   Meters A Meters
• (Choice B)   Minutes B Minutes
• (Choice C)   Meters per minute C Meters per minute
• (Choice D)   Minutes per meter D Minutes per meter

Common mistake: Forgetting to include units, or using incorrect units

Another common mistake: using phrases that refer to “over an interval of time” rather than “at a point in time”, solving problems that involve instantaneous rate of change.

Carlos has taken an initial dose of a prescription medication. The amount of medication, in milligrams, in Carlos's bloodstream after t ‍   hours is given by the following function:

M ( t ) = 20 ⋅ e − 0.8 t ‍  

What is the instantaneous rate of change of the remaining amount of medication after 1 ‍   hour?

• (Choice A)   4.8 ‍   dollars A 4.8 ‍   dollars
• (Choice B)   4.8 ‍   dollars per pound B 4.8 ‍   dollars per pound
• (Choice C)   61 ‍   dollars C 61 ‍   dollars
• (Choice D)   61 ‍   dollars per pound D 61 ‍   dollars per pound

Common mistake: Evaluating the original function rather than the derivative

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Rate of Change

Rate of change is an important mathematical concept. But what exactly is the rate of change? How can we find it? What can it teach us about math? Let's find out:

Rate of change, explained

A rate of change is simply how fast something is changing in relation to something else. We can express this ratio as:

Rate of change = change in y/change in x

For example, what happens if we plant one apple seed? Eventually, we would get an apple tree. This apple tree would produce even more apples as a result. Let's assume that we figured out that one apple seed leads to 200 apples within 5 years. We now know our rate of change.

The above is an example of a positive rate of change. Planting more seeds leads to more apples. But what happens if we cut down a tree instead? This would lead to a net loss of apples in the future. In other words, it is an example of a negative rate of change.

What happens if we neither plant new trees nor cut them down? The number of produced apples would stay the same each year. In other words, this is an example of zero rate of change.

Visualizing rate of change

It's easier to understand how rate of change works if we visualize it. Here's what positive rate of change looks like when we graph it:

As we can see, x and y have a positive relationship. When the value of x increases, so too does the value of y. The same can be said in reverse.

What about negative rate of change? Take a look:

As we can see, the value of y decreases when the value of x increases. This is why we see a characteristic downward slope.

Finally, we have zero rate of change:

Nothing is changing in this example, so the values of x and y stay exactly the same. This is why we see a characteristic straight line.

A real-world example of rate of change

One of the most ubiquitous examples of rate of change is the relationship between time and distance . The longer we travel, the further we get. If we're traveling at a constant speed, we can use this rate of change to determine how far we will travel.

Let's use our knowledge to solve a real-world problem. We can assume the following:

• We can drive 80 miles in 2 hours
• We can drive 160 miles in 4 hours
• We can drive 240 miles in 6 hours

Let's start by establishing our rate of change. We know that the formula for rate of change is the following:

Rate of change is defined as the change in y divided by the change in x.

160 - 80 4 - 2 = 80 2 = 40 1

Now we know that our rate of change is 40 1 or 40 . This tells us that our car is traveling at 40 miles per hour. We can graph this rate of change by plugging in our known values and connecting the dots:

Using the rate of change formula

You can use the rate of change formula to make many important calculations.

• The increase of current in amperes is based on the increase in voltage
• Spotting security momentum and other trends in the financial world
• How much work you can get done within a specific amount of time
• How many people are required to get a specific job done in a certain time
• Projecting future earnings or saving your money

With all that said, you will notice that many equations related to rates of change involve time. This is because time is one of the most important constants in our universe.

Topics related to the Rate of Change

Rate-Time-Distance Problems

Rates and Ratios

Flashcards covering the Rate of Change

Algebra 1 Flashcards

College Algebra Flashcards

Practice tests covering the Rate of Change

Algebra 1 Diagnostic Tests

College Algebra Diagnostic Tests

Pair your student with a tutor who understands rate of change

If your student needs help understanding rate of change, a tutor can provide them with an encouraging, patient 1-on-1 environment. Students can learn about concepts in new ways that may be more conducive to their learning styles. For example, a verbal learner might have trouble understanding rate of change because their classroom teacher is displaying information primarily through diagrams. A tutor can switch things up, allowing verbal learners to talk about concepts instead. This is just one example of how a tutor can personalize your student's learning. To learn more about the possibilities, reach out to our Educational Directors today. Varsity Tutors will pair your student with a great tutor.

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3.4 Derivatives as Rates of Change

Learning objectives.

• 3.4.1 Determine a new value of a quantity from the old value and the amount of change.
• 3.4.2 Calculate the average rate of change and explain how it differs from the instantaneous rate of change.
• 3.4.3 Apply rates of change to displacement, velocity, and acceleration of an object moving along a straight line.
• 3.4.4 Predict the future population from the present value and the population growth rate.
• 3.4.5 Use derivatives to calculate marginal cost and revenue in a business situation.

In this section we look at some applications of the derivative by focusing on the interpretation of the derivative as the rate of change of a function. These applications include acceleration and velocity in physics, population growth rates in biology, and marginal functions in economics.

Amount of Change Formula

One application for derivatives is to estimate an unknown value of a function at a point by using a known value of a function at some given point together with its rate of change at the given point. If f ( x ) f ( x ) is a function defined on an interval [ a , a + h ] , [ a , a + h ] , then the amount of change of f ( x ) f ( x ) over the interval is the change in the y y values of the function over that interval and is given by

The average rate of change of the function f f over that same interval is the ratio of the amount of change over that interval to the corresponding change in the x x values. It is given by

As we already know, the instantaneous rate of change of f ( x ) f ( x ) at a a is its derivative

For small enough values of h , f ′ ( a ) ≈ f ( a + h ) − f ( a ) h . h , f ′ ( a ) ≈ f ( a + h ) − f ( a ) h . We can then solve for f ( a + h ) f ( a + h ) to get the amount of change formula:

We can use this formula if we know only f ( a ) f ( a ) and f ′ ( a ) f ′ ( a ) and wish to estimate the value of f ( a + h ) . f ( a + h ) . For example, we may use the current population of a city and the rate at which it is growing to estimate its population in the near future. As we can see in Figure 3.22 , we are approximating f ( a + h ) f ( a + h ) by the y y coordinate at a + h a + h on the line tangent to f ( x ) f ( x ) at x = a . x = a . Observe that the accuracy of this estimate depends on the value of h h as well as the value of f ′ ( a ) . f ′ ( a ) .

Here is an interesting demonstration of rate of change.

Example 3.33

Estimating the value of a function.

If f ( 3 ) = 2 f ( 3 ) = 2 and f ′ ( 3 ) = 5 , f ′ ( 3 ) = 5 , estimate f ( 3.2 ) . f ( 3.2 ) .

Begin by finding h . h . We have h = 3.2 − 3 = 0.2 . h = 3.2 − 3 = 0.2 . Thus,

Checkpoint 3.21

Given f ( 10 ) = −5 f ( 10 ) = −5 and f ′ ( 10 ) = 6 , f ′ ( 10 ) = 6 , estimate f ( 10.1 ) . f ( 10.1 ) .

Motion along a Line

Another use for the derivative is to analyze motion along a line. We have described velocity as the rate of change of position. If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. It is also important to introduce the idea of speed , which is the magnitude of velocity. Thus, we can state the following mathematical definitions.

Let s ( t ) s ( t ) be a function giving the position of an object at time t . t .

The velocity of the object at time t t is given by v ( t ) = s ′ ( t ) . v ( t ) = s ′ ( t ) .

The speed of the object at time t t is given by | v ( t ) | . | v ( t ) | .

The acceleration of the object at t t is given by a ( t ) = v ′ ( t ) = s ″ ( t ) . a ( t ) = v ′ ( t ) = s ″ ( t ) .

Example 3.34

Comparing instantaneous velocity and average velocity.

A ball is dropped from a height of 64 feet. Its height above ground (in feet) t t seconds later is given by s ( t ) = −16 t 2 + 64 . s ( t ) = −16 t 2 + 64 .

• What is the instantaneous velocity of the ball when it hits the ground?
• What is the average velocity during its fall?

The first thing to do is determine how long it takes the ball to reach the ground. To do this, set s ( t ) = 0 . s ( t ) = 0 . Solving −16 t 2 + 64 = 0 , −16 t 2 + 64 = 0 , we get t = 2 , t = 2 , so it take 2 seconds for the ball to reach the ground.

• The instantaneous velocity of the ball as it strikes the ground is v ( 2 ) . v ( 2 ) . Since v ( t ) = s ′ ( t ) = −32 t , v ( t ) = s ′ ( t ) = −32 t , we obtain v ( t ) = −64 ft/s . v ( t ) = −64 ft/s .
• The average velocity of the ball during its fall is v a v e = s ( 2 ) − s ( 0 ) 2 − 0 = 0 − 64 2 = −32 ft/s . v a v e = s ( 2 ) − s ( 0 ) 2 − 0 = 0 − 64 2 = −32 ft/s .

Example 3.35

Interpreting the relationship between v ( t ) v ( t ) and a ( t ) a ( t ).

A particle moves along a coordinate axis in the positive direction to the right. Its position at time t t is given by s ( t ) = t 3 − 4 t + 2 . s ( t ) = t 3 − 4 t + 2 . Find v ( 1 ) v ( 1 ) and a ( 1 ) a ( 1 ) and use these values to answer the following questions.

• Is the particle moving from left to right or from right to left at time t = 1 ? t = 1 ?
• Is the particle speeding up or slowing down at time t = 1 ? t = 1 ?

Begin by finding v ( t ) v ( t ) and a ( t ) . a ( t ) .

v ( t ) = s ′ ( t ) = 3 t 2 - 4 v ( t ) = s ′ ( t ) = 3 t 2 - 4 and a ( t ) = v ′ ( t ) = s ″ ( t ) = 6 t . a ( t ) = v ′ ( t ) = s ″ ( t ) = 6 t .

Evaluating these functions at t = 1 , t = 1 , we obtain v ( 1 ) = −1 v ( 1 ) = −1 and a ( 1 ) = 6 . a ( 1 ) = 6 .

• Because v ( 1 ) < 0 , v ( 1 ) < 0 , the particle is moving from right to left.
• Because v ( 1 ) < 0 v ( 1 ) < 0 and a ( 1 ) > 0 , a ( 1 ) > 0 , velocity and acceleration are acting in opposite directions. In other words, the particle is being accelerated in the direction opposite the direction in which it is traveling, causing | v ( t ) | | v ( t ) | to decrease. The particle is slowing down.

Example 3.36

Position and velocity.

The position of a particle moving along a coordinate axis is given by s ( t ) = t 3 − 9 t 2 + 24 t + 4 , t ≥ 0 . s ( t ) = t 3 − 9 t 2 + 24 t + 4 , t ≥ 0 .

• Find v ( t ) . v ( t ) .
• At what time(s) is the particle at rest?
• On what time intervals is the particle moving from left to right? From right to left?
• Use the information obtained to sketch the path of the particle along a coordinate axis.
• The velocity is the derivative of the position function: v ( t ) = s ′ ( t ) = 3 t 2 − 18 t + 24 . v ( t ) = s ′ ( t ) = 3 t 2 − 18 t + 24 .
• The particle is at rest when v ( t ) = 0 , v ( t ) = 0 , so set 3 t 2 − 18 t + 24 = 0 . 3 t 2 − 18 t + 24 = 0 . Factoring the left-hand side of the equation produces 3 ( t − 2 ) ( t − 4 ) = 0 . 3 ( t − 2 ) ( t − 4 ) = 0 . Solving, we find that the particle is at rest at t = 2 t = 2 and t = 4 . t = 4 .

Checkpoint 3.22

A particle moves along a coordinate axis. Its position at time t t is given by s ( t ) = t 2 − 5 t + 1 . s ( t ) = t 2 − 5 t + 1 . Is the particle moving from right to left or from left to right at time t = 3 ? t = 3 ?

Population Change

In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types of populations, including those as diverse as bacteria colonies and cities. We can use a current population, together with a growth rate, to estimate the size of a population in the future. The population growth rate is the rate of change of a population and consequently can be represented by the derivative of the size of the population.

If P ( t ) P ( t ) is the number of entities present in a population, then the population growth rate of P ( t ) P ( t ) is defined to be P ′ ( t ) . P ′ ( t ) .

Example 3.37

Estimating a population.

The population of a city is tripling every 5 years. If its current population is 10,000, what will be its approximate population 2 years from now?

Let P ( t ) P ( t ) be the population (in thousands) t t years from now. Thus, we know that P ( 0 ) = 10 P ( 0 ) = 10 and based on the information, we anticipate P ( 5 ) = 30 . P ( 5 ) = 30 . Now estimate P ′ ( 0 ) , P ′ ( 0 ) , the current growth rate, using

By applying Equation 3.10 to P ( t ) , P ( t ) , we can estimate the population 2 years from now by writing

thus, in 2 years the population will be 18,000.

Checkpoint 3.23

The current population of a mosquito colony is known to be 3,000; that is, P ( 0 ) = 3,000 . P ( 0 ) = 3,000 . If P ′ ( 0 ) = 100 , P ′ ( 0 ) = 100 , estimate the size of the population in 3 days, where t t is measured in days.

Changes in Cost and Revenue

In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives. The marginal cost is the derivative of the cost function. The marginal revenue is the derivative of the revenue function. The marginal profit is the derivative of the profit function, which is based on the cost function and the revenue function.

If C ( x ) C ( x ) is the cost of producing x items, then the marginal cost M C ( x ) M C ( x ) is M C ( x ) = C ′ ( x ) . M C ( x ) = C ′ ( x ) .

If R ( x ) R ( x ) is the revenue obtained from selling x x items, then the marginal revenue M R ( x ) M R ( x ) is M R ( x ) = R ′ ( x ) . M R ( x ) = R ′ ( x ) .

If P ( x ) = R ( x ) − C ( x ) P ( x ) = R ( x ) − C ( x ) is the profit obtained from selling x items, then the marginal profit M P ( x ) M P ( x ) is defined to be M P ( x ) = P ′ ( x ) = M R ( x ) − M C ( x ) = R ′ ( x ) − C ′ ( x ) . M P ( x ) = P ′ ( x ) = M R ( x ) − M C ( x ) = R ′ ( x ) − C ′ ( x ) .

We can roughly approximate

by choosing an appropriate value for h . h . Since x represents objects, a reasonable and small value for h h is 1. Thus, by substituting h = 1 , h = 1 , we get the approximation M C ( x ) = C ′ ( x ) ≈ C ( x + 1 ) − C ( x ) . M C ( x ) = C ′ ( x ) ≈ C ( x + 1 ) − C ( x ) . Consequently, C ′ ( x ) C ′ ( x ) for a given value of x x can be thought of as the change in cost associated with producing one additional item. In a similar way, M R ( x ) = R ′ ( x ) M R ( x ) = R ′ ( x ) approximates the revenue obtained by selling one additional item, and M P ( x ) = P ′ ( x ) M P ( x ) = P ′ ( x ) approximates the profit obtained by producing and selling one additional item.

Example 3.38

Applying marginal revenue.

Assume that the number of barbeque dinners that can be sold, x , x , can be related to the price charged, p , p , by the equation p ( x ) = 9 − 0.03 x , 0 ≤ x ≤ 300 . p ( x ) = 9 − 0.03 x , 0 ≤ x ≤ 300 .

In this case, the revenue in dollars obtained by selling x x barbeque dinners is given by

Use the marginal revenue function to estimate the revenue obtained from selling the 101st barbeque dinner. Compare this to the actual revenue obtained from the sale of this dinner.

First, find the marginal revenue function: M R ( x ) = R ′ ( x ) = −0.06 x + 9 . M R ( x ) = R ′ ( x ) = −0.06 x + 9 .

Next, use R ′ ( 100 ) R ′ ( 100 ) to approximate R ( 101 ) − R ( 100 ) , R ( 101 ) − R ( 100 ) , the revenue obtained from the sale of the 101st dinner. Since R ′ ( 100 ) = 3 , R ′ ( 100 ) = 3 , the revenue obtained from the sale of the 101st dinner is approximately $3.

The actual revenue obtained from the sale of the 101st dinner is

The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.

Checkpoint 3.24

Suppose that the profit obtained from the sale of x x fish-fry dinners is given by P ( x ) = −0.03 x 2 + 8 x − 50 . P ( x ) = −0.03 x 2 + 8 x − 50 . Use the marginal profit function to estimate the profit from the sale of the 101st fish-fry dinner.

Section 3.4 Exercises

For the following exercises, the given functions represent the position of a particle traveling along a horizontal line; t ≥ 0 t ≥ 0 .

• Find the velocity and acceleration functions.
• Determine the time intervals when the object is slowing down or speeding up.

s ( t ) = 2 t 3 − 3 t 2 − 12 t + 8 s ( t ) = 2 t 3 − 3 t 2 − 12 t + 8

s ( t ) = 2 t 3 − 15 t 2 + 36 t − 10 s ( t ) = 2 t 3 − 15 t 2 + 36 t − 10

s ( t ) = t 1 + t 2 s ( t ) = t 1 + t 2

A model rocket is fired vertically upward from the ground. The distance s s in feet that the rocket travels from the ground after t t seconds is given by s ( t ) = −16 t 2 + 560 t . s ( t ) = −16 t 2 + 560 t .

• Find the velocity of the rocket 3 seconds after being fired.
• Find the acceleration of the rocket 3 seconds after being fired.

A ball is thrown downward with a speed of 8 ft/s from the top of a 64-foot-tall building. After t seconds, its height above the ground is given by s ( t ) = −16 t 2 − 8 t + 64 . s ( t ) = −16 t 2 − 8 t + 64 .

• Determine how long it takes for the ball to hit the ground.
• Determine the velocity of the ball when it hits the ground.

The position function s ( t ) = t 2 − 3 t − 4 s ( t ) = t 2 − 3 t − 4 represents the position of the back of a car backing out of a driveway and then driving in a straight line, where s s is in feet and t t is in seconds. In this case, s ( t ) = 0 s ( t ) = 0 represents the time at which the back of the car is at the garage door, so s ( 0 ) = −4 s ( 0 ) = −4 is the starting position of the car, 4 feet inside the garage.

• Determine the velocity of the car when s ( t ) = 0 . s ( t ) = 0 .
• Determine the velocity of the car when s ( t ) = 14 . s ( t ) = 14 .

The position of a hummingbird flying along a straight line in t t seconds is given by s ( t ) = 3 t 3 − 7 t s ( t ) = 3 t 3 − 7 t meters.

• Determine the velocity of the bird at t = 1 t = 1 sec.
• Determine the acceleration of the bird at t = 1 t = 1 sec.
• Determine the acceleration of the bird when the velocity equals 0.

A potato is launched vertically upward with an initial velocity of 100 ft/s from a potato gun at the top of an 85-foot-tall building. The position of the potato relative to the ground after t t seconds is given by s ( t ) = −16 t 2 + 100 t + 85 . s ( t ) = −16 t 2 + 100 t + 85 .

• Find the velocity of the potato after 0.5 s 0.5 s and 5.75 s . 5.75 s .
• Find the speed of the potato at 0.5 s and 5.75 s.
• Determine when the potato reaches its maximum height.
• Find the acceleration of the potato at 0.5 s and 1.5 s.
• Determine how long the potato is in the air.
• Determine the velocity of the potato upon hitting the ground.

The position function s ( t ) = t 3 − 8 t s ( t ) = t 3 − 8 t gives the position in miles of a freight train where east is the positive direction and t t is measured in hours.

• Determine the direction the train is traveling when s ( t ) = 0 . s ( t ) = 0 .
• Determine the direction the train is traveling when a ( t ) = 0 . a ( t ) = 0 .
• Determine the time intervals when the train is slowing down or speeding up.

The following graph shows the position y = s ( t ) y = s ( t ) of an object moving along a straight line.

• Use the graph of the position function to determine the time intervals when the velocity is positive, negative, or zero.
• Sketch the graph of the velocity function.
• Use the graph of the velocity function to determine the time intervals when the acceleration is positive, negative, or zero.
• Determine the time intervals when the object is speeding up or slowing down.

The cost function, in dollars, of a company that manufactures food processors is given by C ( x ) = 200 + 7 x + x 2 7 , C ( x ) = 200 + 7 x + x 2 7 , where x x is the number of food processors manufactured.

• Find the marginal cost function.
• Use the marginal cost function to estimate the cost of manufacturing the thirteenth food processor.
• Find the actual cost of manufacturing the thirteenth food processor.

The price p p (in dollars) and the demand x x for a certain digital clock radio is given by the price–demand function p = 10 − 0.001 x . p = 10 − 0.001 x .

• Find the revenue function R ( x ) . R ( x ) .
• Find the marginal revenue function.
• Find the marginal revenue at x = 2000 x = 2000 and 5000 . 5000 .

[T] A profit is earned when revenue exceeds cost. Suppose the profit function for a skateboard manufacturer is given by P ( x ) = 30 x − 0.3 x 2 − 250 , P ( x ) = 30 x − 0.3 x 2 − 250 , where x x is the number of skateboards sold.

• Find the exact profit from the sale of the thirtieth skateboard.
• Find the marginal profit function and use it to estimate the profit from the sale of the thirtieth skateboard.

[T] In general, the profit function is the difference between the revenue and cost functions: P ( x ) = R ( x ) − C ( x ) . P ( x ) = R ( x ) − C ( x ) .

Suppose the price-demand and cost functions for the production of cordless drills is given respectively by p = 143 − 0.03 x p = 143 − 0.03 x and C ( x ) = 75,000 + 65 x , C ( x ) = 75,000 + 65 x , where x x is the number of cordless drills that are sold at a price of p p dollars per drill and C ( x ) C ( x ) is the cost of producing x x cordless drills.

• Find the revenue and marginal revenue functions.
• Find R ′ ( 1000 ) R ′ ( 1000 ) and R ′ ( 4000 ) . R ′ ( 4000 ) . Interpret the results.
• Find the profit and marginal profit functions.
• Find P ′ ( 1000 ) P ′ ( 1000 ) and P ′ ( 4000 ) . P ′ ( 4000 ) . Interpret the results.

A small town in Ohio commissioned an actuarial firm to conduct a study that modeled the rate of change of the town’s population. The study found that the town’s population (measured in thousands of people) can be modeled by the function P ( t ) = − 1 3 t 3 + 64 t + 3000 , P ( t ) = − 1 3 t 3 + 64 t + 3000 , where t t is measured in years.

• Find the rate of change function P ′ ( t ) P ′ ( t ) of the population function.
• Find P ′ ( 1 ) , P ′ ( 2 ) , P ′ ( 3 ) , P ′ ( 1 ) , P ′ ( 2 ) , P ′ ( 3 ) , and P ′ ( 4 ) . P ′ ( 4 ) . Interpret what the results mean for the town.
• Find P ″ ( 1 ) , P ″ ( 2 ) , P ″ ( 3 ) , P ″ ( 1 ) , P ″ ( 2 ) , P ″ ( 3 ) , and P ″ ( 4 ) . P ″ ( 4 ) . Interpret what the results mean for the town’s population.

[T] A culture of bacteria grows in number according to the function N ( t ) = 3000 ( 1 + 4 t t 2 + 100 ) , N ( t ) = 3000 ( 1 + 4 t t 2 + 100 ) , where t t is measured in hours.

• Find the rate of change of the number of bacteria.
• Find N ′ ( 0 ) , N ′ ( 10 ) , N ′ ( 20 ) , N ′ ( 0 ) , N ′ ( 10 ) , N ′ ( 20 ) , and N ′ ( 30 ) . N ′ ( 30 ) .
• Interpret the results in (b).
• Find N ″ ( 0 ) , N ″ ( 10 ) , N ″ ( 20 ) , N ″ ( 0 ) , N ″ ( 10 ) , N ″ ( 20 ) , and N ″ ( 30 ) . N ″ ( 30 ) . Interpret what the answers imply about the bacteria population growth.

The centripetal force of an object of mass m m is given by F ( r ) = m v 2 r , F ( r ) = m v 2 r , where v v is the speed of rotation and r r is the distance from the center of rotation.

• Find the rate of change of centripetal force with respect to the distance from the center of rotation.
• Find the rate of change of centripetal force of an object with mass 1000 kilograms, velocity of 13.89 m/s, and a distance from the center of rotation of 200 meters.

The following questions concern the population (in millions) of London by decade in the 19th century, which is listed in the following table.

• Using a calculator or a computer program, find the best-fit linear function to measure the population.
• Find the derivative of the equation in a. and explain its physical meaning.
• Find the second derivative of the equation and explain its physical meaning.
• Using a calculator or a computer program, find the best-fit quadratic curve through the data.
• Find the derivative of the equation and explain its physical meaning.

For the following exercises, consider an astronaut on a large planet in another galaxy. To learn more about the composition of this planet, the astronaut drops an electronic sensor into a deep trench. The sensor transmits its vertical position every second in relation to the astronaut’s position. The summary of the falling sensor data is displayed in the following table.

• Using a calculator or computer program, find the best-fit quadratic curve to the data.
• Find the derivative of the position function and explain its physical meaning.
• Find the second derivative of the position function and explain its physical meaning.
• Using a calculator or computer program, find the best-fit cubic curve to the data.
• Using the result from c. explain why a cubic function is not a good choice for this problem.

The following problems deal with the Holling type I, II, and III equations. These equations describe the ecological event of growth of a predator population given the amount of prey available for consumption.

[T] The Holling type I equation is described by f ( x ) = a x , f ( x ) = a x , where x x is the amount of prey available and a > 0 a > 0 is the rate at which the predator meets the prey for consumption.

• Graph the Holling type I equation, given a = 0.5 . a = 0.5 .
• Determine the first derivative of the Holling type I equation and explain physically what the derivative implies.
• Determine the second derivative of the Holling type I equation and explain physically what the derivative implies.
• Using the interpretations from b. and c. explain why the Holling type I equation may not be realistic.

[T] The Holling type II equation is described by f ( x ) = a x n + x , f ( x ) = a x n + x , where x x is the amount of prey available and a > 0 a > 0 is the maximum consumption rate of the predator.

• Graph the Holling type II equation given a = 0.5 a = 0.5 and n = 5 . n = 5 . What are the differences between the Holling type I and II equations?
• Take the first derivative of the Holling type II equation and interpret the physical meaning of the derivative.
• Show that f ( n ) = 1 2 a f ( n ) = 1 2 a and interpret the meaning of the parameter n . n .
• Find and interpret the meaning of the second derivative. What makes the Holling type II function more realistic than the Holling type I function?

[T] The Holling type III equation is described by f ( x ) = a x 2 n 2 + x 2 , f ( x ) = a x 2 n 2 + x 2 , where x x is the amount of prey available and a > 0 a > 0 is the maximum consumption rate of the predator.

• Graph the Holling type III equation given a = 0.5 a = 0.5 and n = 5 . n = 5 . What are the differences between the Holling type II and III equations?
• Take the first derivative of the Holling type III equation and interpret the physical meaning of the derivative.
• Find and interpret the meaning of the second derivative (it may help to graph the second derivative).
• What additional ecological phenomena does the Holling type III function describe compared with the Holling type II function?

[T] The populations of the snowshoe hare (in thousands) and the lynx (in hundreds) collected over 7 years from 1937 to 1943 are shown in the following table. The snowshoe hare is the primary prey of the lynx.

• Graph the data points and determine which Holling-type function fits the data best.
• Using the meanings of the parameters a a and n , n , determine values for those parameters by examining a graph of the data. Recall that n n measures what prey value results in the half-maximum of the predator value.
• Plot the resulting Holling-type I, II, and III functions on top of the data. Was the result from part a. correct?

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 1
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-1/pages/3-4-derivatives-as-rates-of-change

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Key Questions

Explanation:.

Slope is the ratio of the vertical and horizontal changes between two points on a surface or a line. The vertical change between two points is called the rise, and the horizontal change is called the run. The slope equals the rise divided by the run: . This simple equation is called the slope formula.

If #y = f(x+h) = 3 (x + h)^ 2# , (Just plug x + h in for x). So, you get this:

The instantaneous rate of change, or derivative, can be written as dy/dx, and it is a function that tells you the instantaneous rate of change at any point.

#y' = f'(x+h) = (d/(dx)) (3*(x)^2) = 6x * 1 = 6x#

. For example, if x = 1, then the instantaneous rate of change is 6.

Rate of Change Formula helps us to calculate the slope of a line if the coordinates of the points on the line are given. ... If coordinates of any two points of a line are given, then the rate of change is the ratio of the change in the y-coordinates to the change in the x-coordinates.

Hope this helps.

Average Rate of Change

The average rate of change of a function #f(x)# on an interval #[a,b]# can be found by

#("Average Rate of Change")={f(b)-f(a)}/{b-a}#

Find the average rate of change of #f(x)=x^2+3x# on #[1,3]# .

#f(3)=(3)^2+3(3)=18#

#f(1)=(1)^2+3(1)=4#

#("Average Rate of Change")={f(3)-f(1)}/{3-1}={18-4}/{2}=14/2=7#

I hope that this was helpful.

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1.3: Rates of Change and Behavior of Graphs

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Learning Objectives

• Find the average rate of change of a function.
• Use a graph to determine where a function is increasing, decreasing, or constant.
• Use a graph to locate local maxima and local minima.
• Use a graph to locate the absolute maximum and absolute minimum.

Gasoline costs have experienced some wild fluctuations over the last several decades. Table \(\PageIndex{1}\) lists the average cost, in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year.

If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to $3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year. In this section, we will investigate changes such as these.

Finding the Average Rate of Change of a Function

The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in Table \(\PageIndex{1}\) did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of change over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value.

\[\begin{align*} \text{Average rate of change}&=\dfrac{\text{Change in output}}{\text{Change in input}} \\[4pt] &=\dfrac{\Delta y}{\Delta x}\\[4pt] &=\dfrac{y_2-y_1}{x_2-x_1}\\[4pt] &=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\end{align*} \label{1.3.1}\]

The Greek letter \(\Delta\) (delta) signifies the change in a quantity; we read the ratio as “delta-\(y\) over delta-\(x\)” or “the change in \(y\) divided by the change in \(x\).” Occasionally we write \(\Delta f\) instead of \(\Delta y\), which still represents the change in the function’s output value resulting from a change to its input value. It does not mean we are changing the function into some other function.

In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7 years, the average rate of change was

\[\dfrac{\Delta y}{\Delta x}=\dfrac{$1.37}{7 \text{years}}\approx \text{0.196 dollars per year.} \label{1.3.2}\]

On average, the price of gas increased by about 19.6¢ each year. Other examples of rates of change include:

• A population of rats increasing by 40 rats per week
• A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes)
• A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon)
• The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage
• The amount of money in a college account decreasing by $4,000 per quarter

Definition: Rate of Change

A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are “output units per input units.”

The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.

\[\dfrac{\Delta y}{\Delta x}=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\]

Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values \(x_1\) and \(x_2\).

• Calculate the difference \(y_2−y_1=\Delta y\).
• Calculate the difference \(x_2−x_1=\Delta x\).
• Find the ratio \(\dfrac{\Delta y}{\Delta x}\).

Example \(\PageIndex{1}\): Computing an Average Rate of Change

Using the data in Table \(\PageIndex{1}\), find the average rate of change of the price of gasoline between 2007 and 2009.

In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is

\[\begin{align*} \dfrac{\Delta y}{\Delta x}&=\dfrac{y_2−y_1}{x_2−x_1} \\[4pt] &=\dfrac{$2.41−$2.84}{2009−2007} \\[4pt] &=\dfrac{−$0.43}{2 \text{ years}} \\[4pt] &=−$0.22 \text{ per year} \end{align*}\]

Note that a decrease is expressed by a negative change or “negative increase.” A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases.

Exercise \(\PageIndex{1}\)

Using the data in Table \(\PageIndex{1}\), find the average rate of change between 2005 and 2010.

\(\dfrac{$2.84−$2.315}{5 \text{ years}} =\dfrac{$0.535}{5 \text{ years}} =$0.106 \text{per year.}\)

Example \(\PageIndex{2}\): Computing Average Rate of Change from a Graph

Given the function \(g(t)\) shown in Figure \(\PageIndex{1}\), find the average rate of change on the interval \([−1,2]\).

At \(t=−1\), Figure \(\PageIndex{2}\) shows \(g(−1)=4\). At \(t=2\),the graph shows \(g(2)=1\).

The horizontal change \(\Delta t=3\) is shown by the red arrow, and the vertical change \(\Delta g(t)=−3\) is shown by the turquoise arrow. The output changes by –3 while the input changes by 3, giving an average rate of change of

\[\dfrac{1−4}{2−(−1)}=\dfrac{−3}{3}=−1\]

Note that the order we choose is very important. If, for example, we use \(\dfrac{y_2−y_1}{x_1−x_2}\), we will not get the correct answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as \((x_1,y_1)\) and \((x_2,y_2)\).

Example \(\PageIndex{3}\): Computing Average Rate of Change from a Table

After picking up a friend who lives 10 miles away, Anna records her distance from home over time. The values are shown in Table \(\PageIndex{2}\). Find her average speed over the first 6 hours.

Here, the average speed is the average rate of change. She traveled 292 miles in 6 hours, for an average speed of

\[\begin{align*}\dfrac{292−10}{6−0}&=\dfrac{282}{6}\\[4pt] &= 47\end{align*}\]

The average speed is about 47 miles per hour.

Because the speed is not constant, the average speed depends on the interval chosen. For the interval \([2,3]\), the average speed is 63 miles per hour.

Example \(\PageIndex{4}\): Computing Average Rate of Change for a Function Expressed as a Formula

Compute the average rate of change of \(f(x)=x^2−\frac{1}{x}\) on the interval \([2, 4]\).

We can start by computing the function values at each endpoint of the interval.

\[\begin{align*}f(2)&=2^2−\frac{1}{2} f(4)&=4^2−\frac{1}{4} \\[4pt] &=4−\frac{1}{2} &=16−\frac{1}{4} \\[4pt] &=72 &=\frac{63}{4}\end{align*}\]

Now we compute the average rate of change.

\[\begin{align*}\text{Average rate of change} &=\dfrac{f(4)−f(2)}{4−2} \\[4pt] &=\dfrac{\frac{63}{4}-\frac{7}{2}}{4-2} \\[4pt] &=\dfrac{\frac{49}{4}}{2} \\[4pt] &= \dfrac{49}{8}\end{align*}\]

Exercise \(\PageIndex{2}\)

Find the average rate of change of \(f(x)=x−2\sqrt{x}\) on the interval \([1, 9]\).

\(\frac{1}{2}\)

Example \(\PageIndex{5}\): Finding the Average Rate of Change of a Force

The electrostatic force \(F\), measured in newtons, between two charged particles can be related to the distance between the particles \(d\),in centimeters, by the formula \(F(d)=\frac{2}{d^2}\). Find the average rate of change of force if the distance between the particles is increased from 2 cm to 6 cm.

We are computing the average rate of change of \(F(d)=\dfrac{2}{d^2}\) on the interval \([2,6]\).

\[\begin{align*}\text{Average rate of change }&=\dfrac{F(6)−F(2)}{6−2} \\[4pt] &=\dfrac{\frac{2}{6^2}-\frac{2}{2^2}}{6-2} & \text{Simplify} \\[4pt] &=\dfrac{\frac{2}{36}-\frac{2}{4}}{4} \\[4pt] &=\dfrac{-\frac{16}{36}}{4} & \text{Combine numerator terms.} \\[4pt] &=−\dfrac{1}{9} & \text{Simplify}\end{align*}\]

The average rate of change is \(−\frac{1}{9}\) newton per centimeter.

Example \(\PageIndex{6}\): Finding an Average Rate of Change as an Expression

Find the average rate of change of \(g(t)=t^2+3t+1\) on the interval \([0, a]\). The answer will be an expression involving \(a\).

We use the average rate of change formula.

\(\begin{align*}\text{Average rate of change} &=\dfrac{g(a)−g(0)}{a−0} & \text{Evaluate.} \\[4pt] &=\dfrac{(a^2+3a+1)−(0^2+3(0)+1)}{a−0} & \text{Simplify.} \\[4pt] &=\dfrac{a^2+3a+1−1}{a} & \text{Simplify and factor.}\\[4pt] &= \dfrac{a(a+3)}{a} & \text{Divide by the common factor a.}\\[4pt] &= a+3 \end{align*}\)

Exercise \(\PageIndex{3}\)

Find the average rate of change of \(f(x)=x^2+2x−8\) on the interval \([5, a]\).

Using a Graph to Determine Where a Function is Increasing, Decreasing, or Constant

As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure \(\PageIndex{3}\) shows examples of increasing and decreasing intervals on a function.

While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a local maximum . If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is called a local minimum . The plural form is “local minima.” Together, local maxima and minima are called local extrema , or local extreme values, of the function. (The singular form is “extremum.”) Often, the term local is replaced by the term relative. In this text, we will use the term local.

Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain.

For the function whose graph is shown in Figure \(\PageIndex{4}\), the local maximum is 16, and it occurs at \(x=−2\). The local minimum is −16 and it occurs at \(x=2\).

To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure \(\PageIndex{5}\) illustrates these ideas for a local maximum.

These observations lead us to a formal definition of local extrema.

Local Minima and Local Maxima

• A function \(f\) is an increasing function on an open interval if \(f(b)>f(a)\) for every \(a\), \(b\) interval where \(b>a\).
• A function \(f\) is a decreasing function on an open interval if \(f(b)<f(a)\) for every \(a\), \(b\) interval where \(b>a\).

A function \(f\) has a local maximum at a point \(b\) in an open interval \((a,c)\) if \(f(b)\) is greater than or equal to \(f(x)\) for every point \(x\) (\(x\) does not equal \(b\)) in the interval. Likewise, \(f\) has a local minimum at a point \(b\) in \((a,c)\) if \(f(b)\) is less than or equal to \(f(x)\) for every \(x\) (\(x\) does not equal \(b\)) in the interval.

Example \(\PageIndex{7}\) Finding Increasing and Decreasing Intervals on a Graph

Given the function \(p(t)\) in Figure \(\PageIndex{6}\), identify the intervals on which the function appears to be increasing.

We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from \(t=1\) to \(t=3\) and from \(t=4\) on.

In interval notation, we would say the function appears to be increasing on the interval \((1,3)\) and the interval \((4,\infty)\).

Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at \(t=1\), \(t=3\), and \(t=4\). These points are the local extrema (two minima and a maximum).

Example \(\PageIndex{8}\): Finding Local Extrema from a Graph

Graph the function \(f(x)=\frac{2}{x}+\frac{x}{3}\). Then use the graph to estimate the local extrema of the function and to determine the intervals on which the function is increasing.

Using technology, we find that the graph of the function looks like that in Figure \(\PageIndex{7}\). It appears there is a low point, or local minimum, between \(x=2\) and \(x=3\), and a mirror-image high point, or local maximum, somewhere between \(x=−3\) and \(x=−2\)

Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure \(\PageIndex{8}\) provides screen images from two different technologies, showing the estimate for the local maximum and minimum.

Based on these estimates, the function is increasing on the interval \((−\infty,−2.449)\) and \((2.449,\infty)\). Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing approximation algorithms used by each. (The exact location of the extrema is at \(\pm\sqrt{6}\), but determining this requires calculus.)

Exercise \(\PageIndex{8}\)

Graph the function \(f(x)=x^3−6x^2−15x+20\) to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing.

The local maximum appears to occur at \((−1,28)\), and the local minimum occurs at \((5,−80)\). The function is increasing on \((−\infty,−1)\cup(5,\infty)\) and decreasing on \((−1,5)\).

Graph of a polynomial with a local maximum at (-1, 28) and local minimum at (5, -80).

Example \(\PageIndex{9}\): Finding Local Maxima and Minima from a Graph

For the function f whose graph is shown in Figure \(\PageIndex{9}\), find all local maxima and minima.

Observe the graph of \(f\). The graph attains a local maximum at \(x=1\) because it is the highest point in an open interval around \(x=1\).The local maximum is the y-coordinate at \(x=1\), which is 2.

The graph attains a local minimum at \(x=−1\) because it is the lowest point in an open interval around \(x=−1\). The local minimum is the y-coordinate at \(x=−1\), which is −2.

Analyzing the Toolkit Functions for Increasing or Decreasing Intervals

We will now return to our toolkit functions and discuss their graphical behavior in Figure \(\PageIndex{10}\), Figure \(\PageIndex{11}\), and Figure \(\PageIndex{12}\).

Use A Graph to Locate the Absolute Maximum and Absolute Minimum

There is a difference between locating the highest and lowest points on a graph in a region around an open interval (locally) and locating the highest and lowest points on the graph for the entire domain. The y-coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum , respectively. To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function (Figure \(\PageIndex{13}\)).

Not every function has an absolute maximum or minimum value. The toolkit function \(f(x)=x^3\) is one such function.

Absolute Maxima and Minima

• The absolute maximum of \(f\) at \(x=c\) is \(f(c)\) where \(f(c)≥f(x)\) for all \(x\) in the domain of \(f\).
• The absolute minimum of \(f\) at \(x=d\) is \(f(d)\) where \(f(d)≤f(x)\) for all \(x\) in the domain of \(f\).

Example \(\PageIndex{10}\): Finding Absolute Maxima and Minima from a Graph

For the function f shown in Figure \(\PageIndex{14}\), find all absolute maxima and minima.

Observe the graph of \(f\). The graph attains an absolute maximum in two locations, \(x=−2\) and \(x=2\), because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y-coordinate at \(x=−2\) and \(x=2\), which is 16.

The graph attains an absolute minimum at x=3, because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y-coordinate at x=3,which is−10.

Key Equations

• Average rate of change: \(\dfrac{\Delta y}{\Delta x}=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\)

Key Concepts

• A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change is determined using only the beginning and ending data. See Example.
• Identifying points that mark the interval on a graph can be used to find the average rate of change. See Example.
• Comparing pairs of input and output values in a table can also be used to find the average rate of change. See Example.
• An average rate of change can also be computed by determining the function values at the endpoints of an interval described by a formula. See Example and Example.
• The average rate of change can sometimes be determined as an expression. See Example.
• A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See Example.
• A local maximum is where a function changes from increasing to decreasing and has an output value larger (more positive or less negative) than output values at neighboring input values.
• A local minimum is where the function changes from decreasing to increasing (as the input increases) and has an output value smaller (more negative or less positive) than output values at neighboring input values.
• Minima and maxima are also called extrema.
• We can find local extrema from a graph. See Example and Example.
• The highest and lowest points on a graph indicate the maxima and minima. See Example.

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