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Radical expressions and functions are commonly found in various fields of mathematics. This lesson will explore the use of radical functions and how they might be applied to a variety of everyday real-life situations, even just watching TV!

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

How to Calculate the Speed of a Car?

After becoming a subscriber of a streaming platform, Magdalena became interested in a particular crime series. While watching an episode, she found out that after an accident, police can determine the speed of a car before the driver started braking.

The speed of a car can be estimated by using the following formula.

v_{1} (d) = 30 μ d

In this formula,

v_{1} (d)

represents the speed in miles per hour,

μ

is the coefficient of friction between the road surface and the tires, and the braking distance

d

— the length of skid marks on the road — is measured in feet. Assume that

μ = 0.6 .

In another episode, criminals organized illegal drag races.

When crossing the finish line, the speed of a race car with a total weight of

3000

pounds can be modeled by the following function.

v_{2} (p) = 8 3 8 p

In this formula,

p

is the power of the engine of the vehicle in horsepower and

v_{2} (p)

is measured in miles per hour. Because both radical functions measure speed and involve roots, Magdalena does not see any difference between them.

a Determine the domains

D_{1}

and

D_{2}

of the functions.

b Compare the average rate of change of the functions over the interval

[0, 216] .

c Graph the functions. What is their end behavior?

Example

Kleiber's Law

Magdalena shared her streaming account with Kriz so that they can watch a new nature documentary. The documentary is about the work of a Swiss agricultural biologist, Max Kleiber. He researched how the mass of an animal influences the animal's metabolic rate.

Metabolic Rate
The rate at which a person or an animal burns calories to maintain its body weight.

Kleiber's Law states that an animal's metabolic rate

R (w)

can be modeled by the following radical function.

R (w) = 70 4 w^{3}

In this function,

w

is the mass of the animal in kilograms and

R (w)

is measured in kilocalories per day.

a Kriz has a couple of pets. Help Kriz calculate their and their pets' metabolic rates. Each mass is given in the table. Round to

2

decimals if needed and write the metabolic rates in increasing order.

Mammal	Mass
Mouse	$20$ g
Kriz	$81$ kg
Horse	$625$ kg

b Consider two animals with weights

w_{1}

and

w_{2}

such that the first animal metabolizes

64

times more calories per day than the second animal. How many times heavier than the second animal is the first animal?

Hint

a Substitute each mass into the given formula. Use the

n^{th}

Roots of

n^{th}

Powers Property to evaluate the metabolic rate.

b Write the equation for the metabolic rate of a heavier animal. Substitute the formula stated in Kleiber's Law to solve for the weight.

Solution

a The metabolic rate of Kriz and their animals can be calculated using the function stated in Kleiber's Law. Notice that not all masses are given in kilograms. The mass of a mouse can be written in kilograms by using the conversion factor

1 g = 0.001 kg .

w_{mouse} = 20 g = 0.02 kg

Now, substitute

w = 0.02

into

R (w)

to find the metabolic rate of a mouse.

R (w) = 70 4 w^{3}

Substitute

$w = 0.02$

R (0.02) = 70 4 0.02^{3}

▼

Evaluate right-hand side

CalcPow

Calculate power

R (0.02) = 70 4 0.000008

UseCalc

Use a calculator

R (0.02) = 3.722807 \dots

RoundDec

Round to $2$ decimal place(s)

R (0.02) \approx 3.72

Next, Kriz's metabolic rate will be calculated. To do so, substitute

w = 81

into

R (w) .

The

n^{th}

Roots of

n^{th}

Powers Property will be used to evaluate the rate.

R (w) = 70 4 w^{3}

Substitute

$w = 81$

R (81) = 70 4 81^{3}

▼

Evaluate right-hand side

WritePow

Write as a power

R (81) = 70 4 (3^{4})^{3}

$(a^{m})^{n} = (a^{n})^{m}$

R (81) = 70 4 (3^{3})^{4}

$4 a^{4} = ∣ a ∣$

R (81) = 70 ∣ ∣ ∣ 3^{3} ∣ ∣ ∣

CalcPow

Calculate power

R (81) = 70 ∣ 27 ∣

AbsPos

$∣ 27 ∣ = 27$

R (81) = 70 (27)

Multiply

R (81) = 1890

The same process can be used to calculate the metabolic rate of the horse.

R (625) = 70 4 625^{3} = 8750

Finally, the results can be summarized in a table.

Mammal	Mass $(kg)$	Metabolic Rate $(kcal / day)$
Mouse	$0.02$	$\approx 3.72$
Kriz	$81$	$1890$
Horse	$625$	$8750$

b It is given that the metabolic rate

R_{1}

of a heavier animal is

64

times the metabolic rate

R_{2} .

R_{1} = 64 R_{2}

To determine the number of times the animal is heavier, the given formula for the metabolic rate will be substituted into the equation.

R_{1} = 64 R_{2} ⇓ 70 4 w_{1}^{3} = 64 (70 4 w_{2}^{3})

Now, solve the obtained equation for

w_{1} .

70 4 w_{1}^{3} = 64 (70 4 w_{2}^{3})

▼

Solve for

w_{1}

Multiply

70 4 w_{1}^{3} = 4480 4 w_{2}^{3}

DivEqn

$LHS / 70 = RHS / 70$

4 w_{1}^{3} = 64 4 w_{2}^{3}

RaiseEqn

$LHS^{4} = RHS^{4}$

(4 w_{1}^{3})^{4} = (64 4 w_{2}^{3})^{4}

PowProdII

$(a b)^{m} = a^{m} b^{m}$

(4 w_{1}^{3})^{4} = 6 4^{4} (4 w_{2}^{3})^{4}

$(n a)^{n} = a$

w_{1}^{3} = 6 4^{4} (w_{2}^{3})

RadicalEqn

$3 LHS = 3 RHS$

3 w_{1}^{3} = 3 6 4^{4} (w_{2}^{3})

RootProd

$3 a \cdot b = 3 a \cdot 3 b$

3 w_{1}^{3} = 3 6 4^{4} \cdot 3 w_{2}^{3}

RootPowToNumber

$3 a^{3} = a$

w_{1} = 3 6 4^{4} \cdot w_{2}

▼

Evaluate right-hand side

RootToPowD

$n a = a^{\frac{1}{n}}$

w_{1} = (6 4^{4})^{\frac{1}{3}} \cdot w_{2}

$(a^{m})^{n} = (a^{n})^{m}$

w_{1} = (6 4^{\frac{1}{3}})^{4} \cdot w_{2}

PowToRootD

$a^{\frac{1}{n}} = n a$

w_{1} = (364)^{4} \cdot w_{2}

CalcRoot

Calculate root

w_{1} = 4^{4} \cdot w_{2}

CalcPowProd

Calculate power and product

w_{1} = 256 w_{2}

This means that an animal that metabolizes

64

times more calories per day is

256

times heavier.

Example

A Ball Thrown Vertically

While watching a basketball game on the streaming platform, Kriz started thinking about the height of a thrown ball and how it relates to how much time passed since the throw.

They once heard that the relationship is given by a quadratic function. However, Kriz wants to check whether this is also the case for a vertical throw. The following radical equation models how much time passed since the vertical throw.

t = \frac{1}{3 2} (v + v^{2} + 64 (s - h))

In this formula,

v

is the vertical velocity in feet per second,

h

is the corresponding height of the ball, and the ball is thrown from a height of

s .

Both

s

and

h

are measured in feet.

a Help Kriz derive the formula for the height of the ball to check if it is modeled by a quadratic function.

b Assume that Kriz throws a ball from a height of

6

feet with an initial vertical velocity of

32

feet per second. What will the height of the ball be after

2

seconds? Calculate the difference between the maximum and the obtained height.

Hint

a Solve the given equation for

h .

Start by isolating the square root. Then, square the equation.

b Use the formula written in Part A. To evaluate the maximum height, find the

x -

coordinate of the vertex of the corresponding parabola.

Solution

a The formula for the height of a ball can be derived by solving the given equation for

h .

To do so, start by isolating the square root.

t = \frac{1}{3 2} (v + v^{2} + 64 (s - h))

▼

Rewrite

MultEqn

$LHS \cdot 32 = RHS \cdot 32$

32 t = 32 (\frac{1}{3 2}) (v + v^{2} + 64 (s - h))

DenomMultFracToNumber

$32 \cdot \frac{a}{3 2} = a$

32 t = 1 (v + v^{2} + 64 (s - h))

IdPropMult

Identity Property of Multiplication

32 t = v + v^{2} + 64 (s - h)

SubEqn

$LHS - v = RHS - v$

32 t - v = v^{2} + 64 (s - h)

RearrangeEqn

Rearrange equation

v^{2} + 64 (s - h) = 32 t - v

Both sides of the equation can now be squared to solve for

h .

v^{2} + 64 (s - h) = 32 t - v

RaiseEqn

$LHS^{2} = RHS^{2}$

(v^{2} + 64 (s - h))^{2} = (32 t - v)^{2}

▼

Solve for

h

PowSqrt

$(a)^{2} = a$

v^{2} + 64 (s - h) = (32 t - v)^{2}

Distr

Distribute $64$

v^{2} + 64 s - 64 h = (32 t - v)^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

v^{2} + 64 s - 64 h = (32 t)^{2} - 2 (32 t) (v) + v^{2}

SubEqn

$LHS - v^{2} = RHS - v^{2}$

64 s - 64 h = (32 t)^{2} - 2 (32 t) (v)

PowProdII

$(a b)^{m} = a^{m} b^{m}$

64 s - 64 h = 3 2^{2} t^{2} - 2 (32 t) (v)

CalcPowProd

Calculate power and product

64 s - 64 h = 1024 t^{2} - 64 v t

SubEqn

$LHS - 64 s = RHS - 64 s$

- 64 h = 1024 t^{2} - 64 v t - 64 s

DivEqn

$LHS / (- 64) = RHS / (- 64)$

h = \frac{1}{- 6 4} (1024 t^{2} - 64 v t - 64 s)

Distr

Distribute $\frac{1}{- 6 4}$

h = - 16 t^{2} + v t + s

It has been calculated that

h = - 16 t^{2} + v t + s .

This means that the height of the ball is indeed modeled by a quadratic function.

b To calculate the ball's height

2

seconds after the throw, the formula derived in Part A will be used. It is given that the ball is thrown from a height of

6

feet with an initial velocity of

32

feet per second. Therefore,

s = 6

and

v = 32

will be substituted into the formula.

h = - 16 t^{2} + v t + s ⇓ h = - 16 t^{2} + 32 t + 6

Now, substitute

t = 2

into the obtained quadratic equation to determine the desired height.

h = - 16 t^{2} + 32 t + 6

Substitute

$t = 2$

h = - 16 (2)^{2} + 32 (2) + 6

▼

Evaluate right-hand side

CalcPow

Calculate power

h = - 16 (4) + 32 (2) + 6

Multiply

h = - 64 + 64 + 6

AddTerms

Add terms

h = 6

After

2

seconds, the ball will be at a height of

6

feet. This is the height at which the ball was initially thrown. Now, the maximum height of the ball will be found. Notice that the function is already written in standard form.

h = - 16 t^{2} + 32 t + 6

To calculate the maximum value of the function, start by determining the

x -

coordinate of the vertex of the corresponding of parabola.

x = - \frac{b}{2 a}

SubstituteII

$a = - 16$ , $b = 32$

x = - \frac{3 2}{2 ( - 1 6 )}

▼

MultPosNeg

$a (- b) = - a \cdot b$

x = - \frac{3 2}{- 3 2}

RemoveNegFracAndDenom

$- \frac{a}{- b} = \frac{a}{b}$

x = \frac{3 2}{3 2}

QuotOne

$\frac{a}{a} = 1$

x = 1

The

x -

coordinate of the vertex is

1 .

The maximum value

h_{max}

can be calculated by substituting

t = 1

into the function rule.

h_{max} = - 16 t^{2} + 32 t + 6

Substitute

$t = 1$

h_{max} = - 16 (1)^{2} + 32 (1) + 6

▼

Evaluate right-hand side

BaseOne

$1^{a} = 1$

h_{max} = - 16 (1) + 32 (1) + 6

IdPropMult

Identity Property of Multiplication

h_{max} = - 16 + 32 + 6

AddTerms

Add terms

h_{max} = 22

Finally, the difference between the maximum height and the height after

2

seconds will be calculated.

h_{max} - h = 22 - 6 = 16

Example

A Quarter Marathon

Magdalena and Kriz see a news story about the quarter marathon they ran last weekend. The distance of the marathon was

10

kilometers. Participants were divided into groups starting at equal intervals.

Magdalena and Kriz had different strategies for how to adjust their running pace to their physical condition. Magdalena's distance covered after

t

minutes can be modeled by the following linear function.

m (t) = \frac{1}{5} t

In this function,

m (t)

is given in kilometers. On the other hand, Kriz's progress after

t

minutes is modeled by a square root function.

k (t) = 2 t - 16

Similarly,

k (t)

is measured in kilometers. For the following questions, assume that the marathon starts at

t = 0 .

a What was the delay between Magdalena's and Kriz's start times?

b At what time(s) did Magdalena and Kriz run side by side? Solve the corresponding system of equations by graphing.

Hint

a Determine at what values of

t

Magdalena and Kriz start the marathon.

b The desired time is given by the values of

t

such that the distances covered by Magdalena and Kriz are equal. Solve the equation by graphing.

Solution

a It is given that the marathon started at

t = 0 .

Notice that when

t

is greater than

0,

the distance covered by Magdalena was also greater than

0 .

t > 0 \Rightarrow \frac{1}{5} t > 0

This means that Magdalena started the race in the first group. In Kriz's case, the distance covered is given by a square root function. Since the distance

k (t) = 2 t - 16

cannot be an imaginary number, this function is defined for values of

t

such that the radicand is non-negative.

2 t - 16 \geq 0 \Leftrightarrow t \geq 8

Because the function is not defined for values of

t

less than

8,

Kriz must not have started the race yet. Moreover, the square root of a positive number is also positive.

t > 8 \Rightarrow 2 t - 16 > 0

This means that Kriz started running

8

minutes after Magdalena, at

t = 8 .

Consequently, the delay was

8

minutes.

b Recall that Magdalena and Kriz ran in the same marathon. Since the delay between their starts was found in Part A, it can be assumed that both given functions share the same timeline

t .

Consider the functions modeling the distance each student ran one more time.

m (t) k (t) = \frac{1}{5} t = 2 t - 16

Any times at which Magdalena and Kriz ran side by side will be given by the values of

t

such that their covered distances are equal.

m (t) = k (t) ⇓ \frac{1}{5} t = 2 t - 16

The combined equation can be solved by graphing. Start with the linear function. To do so, the function rule must be written in slope-intercept form. Then, the slope

m

and

y -

intercept

b

can be easily identified.

y = m t + b ⇓ y = \frac{1}{5} t + 0

Now, the obtained equation can be graphed by plotting its

y -

intercept. Then, the slope will be used to determine another point that satisfies the equation, and the points will be connected with a line.

Next, the second equation $y = 2 t - 16$ will be graphed in the same coordinate plane by using a table of values. Remember to substitute $t \geq 8$ so that the square root can be calculated right away.

$t$	$2 t - 16$	$y = 2 t - 16$
$8$	$2 (8) - 16$	$0 = 0$
$10$	$2 (10) - 16$	$4 = 2$
$16$	$2 (16) - 16$	$16 = 4$
$26$	$2 (26) - 16$	$36 = 6$
$40$	$2 (40) - 16$	$64 = 8$
$58$	$2 (58) - 16$	$100 = 10$

All these points will be plotted and connected with a smooth curve.

According to the graph, Magdalena ran the first

2

kilometers in

10

minutes, while Kriz ran the first

2

kilometers in

2

minutes. They met

10

minutes from the start of the marathon. After this time, Kriz was in front of Magdalena for a while.

t = 10 \Rightarrow m (t) = k (t)

However, Kriz's rhythm was declining while Magdalena's remained constant. Then,

40

minutes after the start of the race, Magdalena caught up with Kriz and from then on, she took the lead until the end of the marathon. Nevertheless, notice that both ran the marathon in

50

minutes.

t = 40 \Rightarrow m (t) = k (t)

It can be concluded that the distances covered by Magdalena and Kriz were equal at

t = 10

and

t = 40 .

This means that they ran side by side

10

minutes and

40

minutes after the start of the marathon.

Example

Period of a Pendulum

A mathematical pendulum is an idealized model of a simple pendulum. It consists of an object suspended from a fixed point with a string. The string cannot be stretched. Consider a pendulum with a period of

0.5

seconds.

The period

T

of a mathematical pendulum, the time it takes to complete one full cycle in seconds, can be approximated for small swings by the following square root function.

T = 2 π \frac{L}{g}

Here,

L

is the length in of the pendulum in feet and

g

is gravity acceleration in feet per second squared. The pendulum that Kriz built has a length of

3

feet.

a Assume that

g

32

feet per second squared and

π \approx 3.14 .

Write the function for the period of Kriz's pendulum depending on the number of feet

x

by which the pendulum is lengthened. Round the constant to

2

decimals if needed.

b Graph the obtained function. Determine the domain and range of the function. Choose the correct answer.

c How does the value of

x

influence the pendulum's period?

Hint

a What is the length of Kriz's pendulum after it is lengthened by

x ?

b Make a table of values to graph the function. Keep in mind that a real square root is defined for a non-negative radicand.

c What is the end behavior of the obtained function?

Solution

a It is given that Kriz's pendulum has a length of

3

feet. Because the pendulum is about to be lengthened by

x,

the total length is given by the following sum.

L = 3 + x

To write the desired function, substitute

3 + x

for

L

into the given function for the period of a pendulum.

T = 2 π \frac{L}{g} ⇓ T = 2 π \frac{3 + x}{g}

Next, substitute

g = 32

and

π \approx 3.14

to simplify the function rule.

T = 2 π \frac{3 + x}{g}

Substitute

$g = 32$

T = 2 π \frac{3 + x}{3 2}

$π \approx 3.14$

T = 2 (3.14) \frac{3 + x}{3 2}

▼

Simplify right-hand side

Multiply

T = 6.28 \cdot \frac{3 + x}{3 2}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

T = 6.28 \cdot \frac{3 + x}{3 2}

$a \cdot \frac{b}{c} = \frac{a}{c} \cdot b$

T = \frac{6 . 2 8}{3 2} \cdot 3 + x

UseCalc

Use a calculator

T = 1.110157 \dots \cdot 3 + x

RoundDec

Round to $2$ decimal place(s)

T \approx 1.11 3 + x

b Before graphing the function obtained in Part A, its domain will be determined. Recall that a real square root is defined only for non-negative radicands.

3 + x \geq 0 ⇕ x \geq - 3

Therefore, the domain of the function is all real numbers greater than or equal to

- 3 .

Next, make a table of values to graph the function. Substitute values of

x

so that the square root can be easily calculated.

$x$	$1.11 3 + x$	$T = 1.11 3 + x$
$- 3$	$1.11 3 + (- 3)$	$0$
$- 2$	$1.11 3 + (- 2)$	$1.11$
$1$	$1.11 3 + 1$	$2.22$
$6$	$1.11 3 + 6$	$3.33$
$13$	$1.11 3 + 13$	$4.44$

Next, the points from the table will be plotted on a coordinate plane and connected with a smooth curve.

Since the square root of a greater number is also greater, the function will increase to infinity. Additionally, the square root of a non-negative number is also non-negative. Consequently, the range of the function consists of numbers greater than or equal to $0 .$ The obtained domain and range correspond to option C.

c Once again, consider the graph of the function drawn in Part B.

It was determined that the range of this function consists of real numbers greater than or equal to

0 .

Also, because the value of the function keeps increasing without bound, the end behavior of the function is up.

As x \to + \infty, T \to + \infty

Moreover, the duration of the period of the pendulum increases more and more slowly as the value of

x

increases. Note that this relationship goes both ways. This means that the longer the period is, the longer the pendulum that creates it needs to be.

Example

Profit of Software Start-ups

In addition to watching TV, Magdalena also loves programming and developing simple applications for personal use in her spare time.

She has recently started thinking about doing a summer internship at one of the local start-up companies. The following radical function estimates the annual profit of the first company in millions of dollars.

P_{1} (t) = \frac{1}{2} 3 t - 2 + 2

In this function,

t

is the number of years after

2022 .

The following graph represents the function that models the profit of the second start-up.

a Choose transformations that are applied to graph the function

P_{1}

starting from the graph of the parent function

y = 3 t .

b Compare the average rates of change

R_{1}

and

R_{2}

of the function

P_{1}

over the intervals

[2, 10]

and

[10, 18],

respectively. Choose the correct answers.

c Compare the average rates of change of the functions

P_{1}

and

P_{2}

over the interval

t = 0

t = 10 .

Read the statements given in the table and choose the correct one.

Option	Statement
A	The average increase in profit of the first company will be greater over the next $10$ years.
B	The average increase in profit of the second company will be greater over the next $10$ years.
C	On average, the profit of the companies increases at the same rate.
D	On average, the annual profit of the second company is about $0.175$ millions of dollars.

Hint

a Take a close look at the function rule of

P_{1} .

What is the difference between vertical and horizontal translation? What type of transformation is applied when the function rule is multiplied by a positive number?

b Use the formula for the average rate of change of a function.

c Similar to in Part B, use the formula for the average rate of change.

Solution

a Consider the parent function and the annual profit function

P_{1} .

All the differences between the function rules will be highlighted.

y P_{1} (t) = \frac{1}{2} 3 t = \frac{1}{2} 3 t - 2 + 2

First, notice that

2

is subtracted from the rule's input

t .

This means that the graph of the function is horizontally translated to the right by

2

units.

Next, notice that the cube root in the function rule is multiplied by a positive number that is less than

1 .

This means that the obtained graph should now be shrunk vertically by a factor of

\frac{1}{2} .

To finish the function rule of

P_{1},

2

is added to

\frac{1}{2} 3 t - 2 .

This implies that the corresponding graph will be vertically translated up

2

units.

Finally, all of the applied transformations will be concluded.

Horizontal translation $2$ units right.
Vertical shrink by a factor of $\frac{1}{2} .$
Vertical translation $2$ units up.

Keep in mind that the order of the transformations may differ. For example, the horizontal translation could be the last transformation.

b The average rate of change of the function

f (x)

over the interval

[x_{1}, x_{2}]

can be calculated by using the following formula.

Average Rate of Change = \frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}}

To find the average rate of change of

P_{1}

over the intervals

[2, 10]

and

[10, 18],

first, the values

P_{1} (2),

P_{1} (10),

and

P_{1} (18)

will be calculated in a table of values. Use a calculator if necessary.

$t$	$\frac{1}{2} 3 t - 2 + 2$	$P_{1} (t)$
$2$	$\frac{1}{2} 3 2 - 2 + 2$	$2$
$10$	$\frac{1}{2} 3 10 - 2 + 2$	$3$
$18$	$\frac{1}{2} 3 18 - 2 + 2$	$\approx 3.26$

Now, substitute values of $t$ and $P_{1} (t)$ into the formula for the average rate of change over both intervals.

Interval $[t_{1}, t_{2}]$	$\frac{P _{1} ( t _{2} ) - P _{1} ( t _{1} )}{t _{2} - t _{1}}$	Substitute and Evaluate
$[2, 10]$	$\frac{P _{1} ( 1 0 ) - P _{1} ( 2 )}{1 0 - 2}$	$\frac{3 - 2}{1 0 - 2} = 0.125$
$[10, 18]$	$\frac{P _{1} ( 1 8 ) - P _{1} ( 1 0 )}{1 8 - 1 0}$	$\frac{3 . 2 6 - 3}{1 8 - 1 0} \approx 0.033$

The average rate of change $R_{1}$ is greater than $R_{2} .$ Also, both values are positive. This means that after $10$ years, the profit of the first company will increase but more slowly.

c Similar to Part B, the formula for the average rate of change of a function will be used. Start by calculating

P_{1} (0)

and

P_{1} (10) .

$t$	$\frac{1}{2} 3 t - 2 + 2$	$P_{1} (t)$
$0$	$\frac{1}{2} 3 0 - 2 + 2$	$\approx 1.37$
$10$	$\frac{1}{2} 3 10 - 2 + 2$	$3$

Next, approximate the values of $P_{2} (0)$ and $P_{2} (10)$ from the graph of the profit $P_{2} .$

It was found that $P_{2} (0)$ is about $1.25$ and $P_{2} (10)$ is equal to $3 .$ Now, both average rates of change over the interval $[0, 10]$ can be determined. Use a calculator if needed.

Profit Function	$\frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}}$	Substitute and Evaluate
$P_{1}$	$\frac{P _{1} ( 1 0 ) - P _{1} ( 0 )}{1 0 - 0}$	$\frac{3 - 1 . 3 7}{1 0 - 0} = 0.163$
$P_{2}$	$\frac{P _{2} ( 1 0 ) - P _{2} ( 0 )}{1 0 - 0}$	$\frac{3 - 1 . 2 5}{1 0 - 0} = 0.175$

The average rate of change of the annual profit of the second company is greater over the next $10$ years. This means that the average increase in profit of the second company is greater. Keep in mind that the average rate of change measures the average change in profit, not the actual profit.

Closure

Comparing Different Ways of Calculating the Speed of a Car

In the given challenge, two different functions that model the speed of a car are given. The first function estimates the speed based on the braking distance, while the second function uses the power of the car to approximate its speed.

v_{1} (d) v_{2} (p) = 30 μ d = 8 3 8 p

In these formulas,

μ = 0.6

is the coefficient of friction,

d

is the braking distance in feet, and

p

is the power in horsepower. Both speeds are measured in miles per hour.

a Find the domains

D_{1}

and

D_{2}

of the functions. Let

R

denote the set of all real numbers.

b Determine the average rate of change of the functions over the interval

[0, 216]

and choose the correct statements.

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c Graph the functions and describe their end behavior.

Hint

a The domain of a radical function depends on the index of the root. Does a negative speed make sense?

b Use the formula for the average rate of change of a function.

c Determine the end behavior of functions based on their graphs.

Solution

a The domains of both functions will be determined one at a time. Start by simplifying the first function rule. To do so, use the fact that

μ,

the coefficient of friction, is equal to

0.6 .

v_{1} (d) = 30 μ d

Substitute

$μ = 0.6$

v_{1} (d) = 30 (0.6) d

Multiply

v_{1} (d) = 18 d

Because the square root of a negative number is imaginary, the input of a square root function must be greater than or equal to

0 .

18 d \geq 0 \Leftrightarrow d \geq 0

This means that the domain of the first function is all real numbers greater than or equal to

0 .

The domain can be written in set-builder notation.

D_{1} = {d ∣ d \geq 0}

Next, consider the second function. The Product Property of Radicals can be used to simplify its function rule.

v_{2} (p) = 8 3 8 p

▼

Simplify right-hand side

RootProd

$3 a \cdot b = 3 a \cdot 3 b$

v_{2} (p) = 838 \cdot 3 p

CalcRoot

Calculate root

v_{2} (p) = 8 (2) \cdot 3 p

Multiply

v_{2} (p) = 16 3 p

The cube root of a negative number is a real number. Mathematically, the domain of this function includes all real numbers

R .

However, here, a negative root would lead to a negative power and, consequently, a negative speed. These do not make sense for a car. Therefore, the domain is all real numbers greater than or equal to

0 .

D_{2} = {p ∣ p \geq 0}

b The average rate of change of the function

f (x)

over the interval

[x_{1}, x_{2}]

can be calculated by using the following formula.

Average Rate of Change = \frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}}

To find the average rate of change of

v_{1}

and

v_{2}

over the interval

[0, 216],

first, the values

v_{1} (0),

v_{1} (216),

v_{2} (0),

and

v_{2} (216)

will be calculated in a table of values. Use a calculator if necessary.

Input	$18 d$	$v_{1} (d)$	$16 3 p$	$v_{2} (p)$
$0$	$18 (0)$	$0$	$1630$	$0$
$216$	$18 (216)$	$\approx 62.35$	$163216$	$96$

Now, both average rates of change over the interval $[0, 216]$ can be calculated by substituting the obtained values.

Function	$\frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}}$	Substitute and Evaluate
$v_{1}$	$\frac{v _{1} ( 2 1 6 ) - v _{1} ( 0 )}{2 1 6 - 0}$	$\frac{6 2 . 3 5 - 0}{2 1 6 - 0} \approx 0.29$
$v_{2}$	$\frac{v _{2} ( 2 1 6 ) - v _{2} ( 0 )}{2 1 6 - 0}$	$\frac{9 6 - 0}{2 1 6 - 0} \approx 0.44$

It can be concluded that, on average, the increase in speed of the second model is greater than the increase in speed of the first model. Also, recall that the average rate of change can be thought of as the slope of the corresponding linear function.

On average, if the input of a function increases by one unit, the value of the function increases by the average rate of change, depending on the interval considered.

This means that the speed of the car increases by about $0.29$ for each meter of braking distance, or skid marks left on the ground by braking, and by about $0.44$ miles for each unit of horsepower. Therefore, options A and D are correct.

c Each function will be graphed one at a time by using a table of values. In Part A, it was determined that the domain of

v_{1}

is all non-negative real numbers. Substitute such numbers into the function rule.

$d$	$18 d$	$v_{1} (d)$
$0$	$18 (0)$	$0$
$2$	$18 (2)$	$6$
$4$	$18 (4)$	$\approx 8.5$
$6$	$18 (6)$	$\approx 10.4$
$8$	$18 (8)$	$12$
$10$	$18 (10)$	$\approx 13.4$

Now, plot the points on a coordinate plane and connect them with a smooth curve.

Based on the graph, the value of the function keeps increasing without bound, so the end behavior of the function is up. Also, the function is not defined for negative values of

d .

As d \to + \infty, v_{1} (d) \to + \infty

Next, just as for the first function, calculate the values of the second function.

$p$	$16 3 p$	$v_{2} (p)$
$- 8$	$16 3 - 8$	$- 32$
$- 6$	$16 3 - 6$	$\approx - 29.1$
$- 4$	$16 3 - 4$	$\approx - 25.4$
$- 2$	$16 3 - 2$	$\approx - 20.2$
$0$	$1630$	$0$
$2$	$1632$	$\approx 20.2$
$4$	$1634$	$\approx 25.4$
$6$	$1636$	$\approx 29.1$
$8$	$1638$	$32$

Similarly, plot the obtained points and connect them with a smooth curve.

By analyzing the left and right ends of the graph, the end behavior of the function can be concluded.

As p \to + \infty, v_{2} (p) \to + \infty, As p \to - \infty, v_{2} (p) \to - \infty

To summarize, the correct options are B, D, and E.

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