problem solving involving operations of functions

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Chapter 11: Functions

11.2 Operations on Functions

In Chapter 5, you solved systems of linear equations through substitution, addition, subtraction, multiplication, and division. A similar process is employed in this topic, where you will add, subtract, multiply, divide, or substitute functions. The notation used for this looks like the following:

Given two functions [latex]f(x)[/latex] and [latex]g(x)[/latex]:

[latex]\begin{array}{clcl} f(x) + g(x)&\text{ is the same as }&(f + g)(x)&\text{ and means the addition of these two functions} \\ f(x) - g(x)&\text{ is the same as }&(f - g)(x)&\text{ and means the subtraction of these two functions} \\ f(x)\cdot g(x)&\text{ is the same as }&(f\cdot g)(x)&\text{ and means the multiplication of these two functions} \\ f(x)\div g(x)&\text{ is the same as }&(f\div g)(x)&\text{ and means the addition of these two functions} \end{array}[/latex]

When encountering questions about operations on functions, you will generally be asked to do two things: combine the equations in some described fashion and to substitute some value to replace the variable in the original equation. These are illustrated in the following examples.

Example 11.2.1

Perform the following operations on [latex]f(x) = 2x^2 - 4[/latex] and [latex]g(x) = x^2 + 4x - 2[/latex].

• [latex]f(x) + g(x)[/latex]Addition yields [latex]2x^2 - 4 + x^2 + 4x - 2[/latex], which simplifies to [latex]3x^2 + 4x - 6[/latex].
• [latex]f(x) - g(x)[/latex]Subtraction yields [latex]2x^2-4-(x^2+4x-2)[/latex], which simplifies to [latex]x^2-4x-2[/latex].
• [latex]f(x)\cdot g(x)[/latex]Multiplication yields [latex](2x^2-4)(x^2+4x-2)[/latex], which simplifies to [latex]2x^4+8x^3-4x^2-16x+8[/latex].
• [latex]f(x)\div g(x)[/latex]Division yields [latex](2x^2-4)\div (x^2+4x-2)[/latex], which cannot be reduced any further.

Often, you are asked to evaluate operations on functions where you must substitute some given value into the combined functions. Consider the following.

Example 11.2.2

Perform the following operations on [latex]f(x) = x^2 - 3[/latex] and [latex]g(x) = 2x^2 + 3x[/latex] and evaluate for the given values.

• [latex]f(2) + g(2)[/latex] [latex][x^2-3]+[2x^2+3x][/latex] [latex][(2)^2-3]+[2(2)^2+3(2)][/latex] [latex]4-3+8+6=15[/latex] [latex]f(2)+g(2)=15[/latex]
• [latex]f(1) - g(3)[/latex] [latex][x^2-3]-[2x^2+3x][/latex] [latex][(1)^2-3]-[2(3)^2+3(3)][/latex] [latex][1-3]-[18+9]=-29[/latex] [latex]f(1)-g(3)=-29[/latex]
• [latex]f(0)\cdot g(2)[/latex] [latex][x^2-3]\cdot [2x^2+3x][/latex] [latex][0^2-3]\cdot [2(2)^2+3(2)][/latex] [latex][-3]\cdot [8+6]=-42[/latex] [latex]f(0)\cdot g(2)=-42[/latex]
• [latex]f(2)\div g(0)[/latex] [latex][x^2-3]\div [2x^2+3x][/latex] [latex][2^2-3]\div [2(0)^2+3(0)][/latex] [latex][1]\div [0]=\text{ undefined}[/latex]

Composite functions are functions that involve substitution of functions, such as [latex]f(x)[/latex] is substituted for the [latex]x[/latex]-value in the [latex]g(x)[/latex] function or the reverse. Which goes where is outlined by the way the equation is written:

[latex]\begin{array}{l} (f \circ g)(x)\text{ means that the }g(x)\text{ function is used to replace the }x\text{-values in the }f(x)\text{ function} \\ (g\circ f)(x)\text{ means that the }f(x)\text{ function is used to replace the }x\text{-values in the }g(x)\text{ function} \end{array}[/latex]

The more conventional way to write these composite functions is:

[latex](f\circ g)(x) = f(g(x))\text{ and }(g\circ f)(x) = g(f(x))[/latex]

Consider the following examples of composite functions.

Example 11.2.3

Given the functions [latex]f(x) = 3x - 5[/latex] and [latex]g(x) = x^2 + 2[/latex], evaluate for:

• [latex](f\circ g)(2)[/latex][latex]\begin{array}{rrl} (f\circ g)(x)&=&f(g(x)) \\ f(g(x))&=&3(x^2+2)-5 \\ f(g(2))&=&3(2^2+2)-5 \\ f(g(2))&=&3(6)-5=13 \end{array}[/latex]
• [latex](g\circ f)(-1)[/latex][latex]\begin{array}{rrl} (g\circ f)(x)&=&g(f(x)) \\ g(f(x))&=&[3x-5]^2+2 \\ g(f(-1))&=&[3(-1)-5]^2+2 \\ g(f(-1))&=&[-8]^2+2 \\ g(f(-1))&=&66 \end{array}[/latex]

Perform the indicated operations.

• [latex]g(a) = a^3 + 5a^2[/latex] [latex]f(a) = 2a + 4[/latex] Find [latex]g(3) + f(3)[/latex]
• [latex]f(x) = -3x^2 + 3x[/latex] [latex]g(x) = 2x + 5[/latex] Find [latex]\dfrac{f(-4)}{g(-4)}[/latex]
• [latex]g(x) = -4x + 1[/latex] [latex]h(x) = -2x - 1[/latex] Find [latex]g(5) + h(5)[/latex]
• [latex]g(x) = 3x + 1[/latex] [latex]f(x) = x^3 + 3x^2[/latex] Find [latex]g(2)\cdot f(2)[/latex]
• [latex]g(t) = t - 3[/latex] [latex]h(t) = -3t^3 + 6t[/latex] Find [latex]g(1) + h(1)[/latex]
• [latex]g(x) = x^2 - 2[/latex] [latex]h(x) = 2x + 5[/latex] Find [latex]g(-6) + h(-6)[/latex]
• [latex]h(n) = 2n - 1[/latex] [latex]g(n) = 3n - 5[/latex] Find [latex]\dfrac{h(0)}{g(0)}[/latex]
• [latex]g(a) = 3a - 2[/latex] [latex]h(a) = 4a - 2[/latex] Find [latex](g + h)(10)[/latex]
• [latex]g(a) = 3a + 3[/latex] [latex]f(a) = 2a - 2[/latex] Find [latex](g + f)(9)[/latex]
• [latex]g(x) = 4x + 3[/latex] [latex]h(x) = x^3 - 2x^2[/latex] Find [latex](g - h)(-1)[/latex]
• [latex]g(x) = x + 3[/latex] [latex]f(x) = -x + 4[/latex] Find [latex](g - f)(3)[/latex]
• [latex]g(x) = x^2 + 2[/latex] [latex]f(x) = 2x + 5[/latex] Find [latex](g - f)(0)[/latex]
• [latex]f(n) = n - 5[/latex] [latex]g(n) = 4n + 2[/latex] Find [latex](f + g)(-8)[/latex]
• [latex]h(t) = t + 5[/latex] [latex]g(t) = 3t - 5[/latex] Find [latex](h\cdot g)(5)[/latex]
• [latex]g(t) = t - 4[/latex] [latex]h(t) = 2t[/latex] Find [latex](g\cdot h)(3t)[/latex]
• [latex]g(n) = n^2 + 5[/latex] [latex]f(n) = 3n + 5[/latex] Find [latex]\dfrac{g(n)}{f(n)}[/latex]
• [latex]g(a) = -2a + 5[/latex] [latex]f(a) = 3a + 5[/latex] Find [latex]\left(\dfrac{g}{f}\right)(a^2)[/latex]
• [latex]h(n) = n^3 + 4n[/latex] [latex]g(n) = 4n + 5[/latex] Find [latex]h(n) + g(n)[/latex]
• [latex]g(n) = n^2 - 4n[/latex] [latex]h(n) = n - 5[/latex] Find [latex]g(n^2)\cdot h(n^2)[/latex]
• [latex]g(n) = n + 5[/latex] [latex]h(n) = 2n - 5[/latex] Find [latex](g\cdot h)(-3n)[/latex]

Solve the following composite functions.

• [latex]f(x) = -4x + 1[/latex] [latex]g(x) = 4x + 3[/latex] Find [latex](f\circ g)(9)[/latex]
• [latex]h(a) = 3a + 3[/latex] [latex]g(a) = a + 1[/latex] Find [latex](h \circ g)(5)[/latex]
• [latex]g(x) = x + 4[/latex] [latex]h(x) = x^2 - 1[/latex] Find [latex](g \circ h)(10)[/latex]
• [latex]f(n) = -4n + 2[/latex] [latex]g(n) = n + 4[/latex] Find [latex](f \circ g)(9)[/latex]
• [latex]g(x) = 2x - 4[/latex] [latex]h(x) = 2x^3 + 4x^2[/latex] Find [latex](g \circ h)(3)[/latex]
• [latex]g(x) = x^2 - 5x[/latex] [latex]h(x) = 4x + 4[/latex] Find [latex](g \circ h)(x)[/latex]
• [latex]f(a) = -2a + 2[/latex] [latex]g(a) = 4a[/latex] Find [latex](f \circ g)(a)[/latex]
• [latex]g(x) = 4x + 4[/latex] [latex]f(x) = x^3 - 1[/latex] Find [latex](g \circ f)(x)[/latex]
• [latex]g(x) = -x + 5[/latex] [latex]f(x) = 2x - 3[/latex] Find [latex](g \circ f)(x)[/latex]
• [latex]f(t) = 4t + 3[/latex] [latex]g(t) = -4t - 2[/latex] Find [latex](f \circ g)(t)[/latex]

Answer Key 11.2

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Operations with Functions

Let us try doing those operations on f(x) and g(x) :

We can add two functions:

(f+g)(x) = f(x) + g(x)

Note: we put the f+g inside () to show they both work on x .

Example: f(x) = 2x+3 and g(x) = x 2

(f+g)(x) = (2x+3) + (x 2 ) = x 2 +2x+3

Sometimes we may need to combine like terms :

Example: v(x) = 5x+1, w(x) = 3x-2

(v+w)(x) = (5x+1) + (3x-2) = 8x-1

The only other thing to worry about is the Domain (the set of numbers that go into the function), but we will talk about that later!

We can also subtract two functions:

(f-g)(x) = f(x) − g(x)

(f-g)(x) = (2x+3) − (x 2 )

We can multiply two functions:

(f·g)(x) = f(x) · g(x)

(f · g)(x) = (2x+3)(x 2 ) = 2x 3 + 3x 2

And we can divide two functions:

(f/g)(x) = f(x) / g(x)

(f/g)(x) = (2x+3)/x 2

Function Composition

It has been easy so far, but now we must consider the Domains of the functions.

The domain is the set of all the values that go into a function.

The function must work for all values we give it, so it is up to us to make sure we get the domain correct!

Example: the domain for √x (the square root of x)

We can't have the square root of a negative number (unless we use imaginary numbers, but we aren't doing that here), so we must exclude negative numbers:

The Domain of √x is all non-negative Real Numbers

On the Number Line it looks like:

Using set-builder notation it is written:

"the set of all x's that are a member of the Real Numbers, such that x is greater than or equal to zero "

Or using interval notation it is:

It is important to get the Domain right, or we will get bad results!

So how do we work out the new domain after doing an operation?

How to Work Out the New Domain

When we do operations on functions, we end up with the restrictions of both .

It is like cooking for friends:

• one can't eat peanuts,
• the other can't eat dairy food.

So what we cook can't have peanuts and also can't have dairy products.

Example: f(x)=√x and g(x)=√(3−x)

The domain for f(x)=√x is from 0 onwards:

The domain for g(x)=√(3−x) is up to and including 3:

So the new domain (after adding or whatever) is from 0 to 3:

If we choose any other value, then one or the other part of the new function won't work.

In other words we want to find where the two domains intersect .

Note: we can put this whole idea into one line using Set Builder Notation :

The same rule applies when we add, subtract, multiply or divide, except divide has one extra rule.

An Extra Rule for Division

There is an extra rule for division:

As well as restricting the domain as above, when we divide :

we must also make sure that g(x) is not equal to zero (so we don't divide by zero ).

Here is an example:

(f/g)(x) = √x / √(3−x)

1. The domain for f(x)=√x is from 0 onwards:

2. The domain for g(x)=√(3−x) is up to and including 3:

3. AND √(3−x) cannot be zero , so x cannot be 3:

So all together we end up with:

• To add, subtract, multiply or divide functions just do as the operation says.
• The domain of the new function will have the restrictions of both functions that made it.
• Divide has the extra rule that the function we are dividing by cannot be zero.

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SAT II Math I : Solving Functions from Word Problems

Study concepts, example questions & explanations for sat ii math i, all sat ii math i resources, example questions, example question #1 : solving functions from word problems.

At Joe's pizzeria a pizza costs $5 with the first topping, and then an additional 75 cents for each additional topping.

Notice that the question describes a linear equation because there is a constant rate of change (the cost per topping). This means we can use slope intercept form to describe the scenario. 

Recall that slope intercept form is

Putting all these steps together we get:

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Operations on functions

This lesson will teach you how to perform operations on functions. Basically, you can add, subtract, multiply, and divide functions. 

Examples showing how to do operations on functions.

How to do addition of functions

( f + g )(x) = f(x) + g(x)

Find f + g if  f(x) = 2x - 7 and g(x) = 5x + 8

(f + g)(x)   =   f(x) + g(x)                  = 2x - 7 + 5x + 8                  = 2x + 5x + - 7 + 8                   = 7x + 1

How to do subtraction of functions

( f - g )(x) = f(x) - g(x)

Find f - g if  f(x) = 3x + 9 and g(x) = 6x - 3

(f - g)(x)   =   f(x) - g(x)                  = 3x + 9  -  (6x - 3)                  = 3x + 9  - 6x - - 3                     = 3x + 9 - 6x + 3                  = 3x - 6x + 9 + 3                 = -3x + 12

Notice that we were able to cancel x + 1 since x + 1 is on top and at the bottom in the rational expression.

Let f and g be functions. Take a look at the following figure to see how we can perform these operations on a function.

Composite function

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Function Composition in Word Problems

Sets of Points Functions at Points Functions into Functions Domains & Decompositions Word Probs Inverse Functions

Oftentimes, we do multiple sequential mathematical operations when we're doing our scratch-work. For instance, we might do a multiplication, and then do a subtraction, using "equals" signs as we go — even though the logic isn't correct.

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We often do this when we're working vertically, such as the following:

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120 x 27 ---- 840 240 ---- 3240 -1400 ---- 1840

Strictly speaking, the above was correct until I did the subtraction; 120×27 (at the beginning) is not equal to 1840 (at the end). I should have done the subtraction as its own step:

120 x 27 3240 ---- -1400 840 ----- 240 1840 ---- 3240

If you've done the first sort of computation and your instructor said that this wasn't right, that you had to show each step separately— Well, your instructor was correct, in the sense that you were putting "equals" signs between things that weren't actually equal. But, if you think about it a different way, you were kind-of composing functions.

How so? If we look at the computations as a sequence of operations to be applied to some given input, then the above could be viewed as f  ( x ) = 27 x and g ( x ) =  x  − 1400 , with the input value being x  = 120 . And the sequence of steps could have been done as a function composition; namely, as ( g  ∘  f  )(120)

Does anyone do function composition in real life?

Composition of functions, as a process at least, is used by people every day in real life. Composition of functions allows us to do many computations in a row; when working on real-life things like taxes, one often does many computations in a row. Thus, many of the steps for filing one's taxes may be viewed as representing the composition of functions. For instance, instructions for some section may say something like this:

Take the value from line 31. Multiply by 0.03. Subtract 10,500 from this value. Write down this amount or, if this amount is less than zero, write down zero.

If we label the original "value from line 31" as x , then the multiplication could be viewed as being f  ( x ) = 0.03 x . The subtracting could be g ( x ) =  x  − 10,500 . Then the last step would be something like h ( x ) = max( x , 0) . The tax computation, from beginning to end, could be viewed as the following composition:

h ( g ( f ( x )))

So composition of functions can be implicit in everyday life. As a result, you may be given "real life" word problems involving composition. For these exercises, you'll need to think about what's going on, the order in which things are being done, and therefore the way in which the modelling functions need to be composed.

What is an example of a word problem that uses function composition?

• You work forty hours a week at a furniture store. You receive a $220 weekly salary, plus a 3% commision on sales over $5000. Assume that you sell enough this week to get the commission. Given the functions f  ( x ) = 0.03 x and g ( x ) =  x − 5000 , which of ( f  ∘  g )( x ) and ( g  ∘  f  )( x ) represents your commission?

Well, ( f  ∘  g )( x ) =  f  ( g ( x )) would mean that I would take my sales x , subtract off the $5000 that didn't get the commission, and then multiply whatever is left by 3% .

On the other hand, ( g  ∘  f  )( x ) =  g ( f  ( x )) would mean that I would take my sales x , multiply by 3% , and then subtract $5000 from the result. Not only is this not how the commission is calculator, this could land me in negative numbers! Would I owe money back to my boss...?

So ( f  ∘  g )( x ) is the compostion that does what I need it to do.

( f ∘ g )( x ) represents my commission.

If you're not sure how the formulas are working, try plugging in numbers that you *can* understand, and pay attention to what you do with those numbers. The formula you need will represent the same process as whatever you did. In the case of the commission formula above, you could test the following sales values (all above the minimum for commissions):

• total sales: $6000
• commission sales: $6000 − $5000 = $1000
• commission: 0.03($1000) = $30
• total sales: $8000
• commission sales: $8000 − $5000 = $3000
• commission: 0.03($3000) = $90
• total sales: $15000
• commission sales: $15000 − $5000 = $10000
• commission: 0.03($10000) = $300

For each sales value, I first subtracted $5000 to see how much was getting a commission. Then I multiplied this amount by 3% to find out how much commission I was getting. By looking at this pattern, I could see that I should apply the "subtract five thousand" formula first, and apply the "multiply by three percent" formula last. This matches f  ( g ( x )) = ( f  ∘  g )( x ) , which confirms my earlier answer.

• Write a function t ( x ) for the total, after taxes, on the purchase amount x . Write another function f  ( x ) for the total, including the delivery fee, on the purchase amount x .
• Calculate and interpret ( f  ∘  t )( x ) and ( t  ∘  f  )( x ) . Which results in a lower cost to you?
• Suppose taxes, by law, are not to be charged on delivery fees (which represent a "service"). Which composite function must then be used?

This sort of calculation actually comes up in "real life", and is used for programming the cash registers. And this is why there is a separate button on the register for delivery fees and why they're not rung up as just another purchase.

( i )  The taxes are 7.5% , so the tax function is given by t ( x ) = 1.075 x

The delivery fee is fixed, so the purchase amount is irrelevant. No matter the total of items and taxes, the fee is always going to be "twenty bucks added at the end".

The fee function is given by f  ( x ) =  x  + 20

( ii ) Composing, I get this:

( f ∘ t )( x ) = f  ( t ( x ))

      = f  (1.075 x )

      = 1.075 x + 20

( t ∘ f  )( x ) = t ( f  ( x ))

      = t ( x + 20)

      = 1.075( x + 20)

      = 1.075 x + 21.50

Then I would pay more using ( t  ∘  f  )( x ) , because I would be paying taxes (from the t ( x ) formula) on the delivery fee (the " +20 " in the formula for f  ( x ) ). I would prefer that the delivery fee be tacked on after the taxes, because:

( f ∘ t )( x ) results in a lower cost to me.

( iii )  If the state is not allowed to collect taxes on delivery fees, then:

The function to use is ( f  ∘  t )( x ) .

• Your computer's screen saver is an expanding circle. The circle starts as a dot in the middle of the screen and expands outward, changing colors as it grows. With a twenty-one inch screen, you have a viewing area with a 10-inch radius (measured from the center diagonally down to a corner). The circle reaches the corners in four seconds. Express the area of the circle (discounting the area cut off by the edges of the viewing area) as a function of time t in seconds.

Since the circle's leading edge covers ten inches in four seconds, the radius is growing at a rate of (10 inches)/(4 seconds) = 2.5 inches per second. Then the equation of the radius r , as a function of time t , is:

r ( t ) = 2.5 t

The formula for the area A of a circle, as a function of the radius r , is given by:

A ( r ) = π r 2

Then the circle's area, as a function of time, is found by plugging the radius equation into the area equation, and simplifying the composition:

A ( t ) = ( A ∘ r )( t ) = A ( r ( t ))

= A (2.5 t )

= π(2.5 t ) 2

= π(6.25 t 2 )

= 6.25π t 2

Then the function they're looking for, stating the area A as a function of the time t , is:

A ( t ) = 6.25 πt 2

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Rational Functions

Solve applied problems involving rational functions.

In Example 2, we shifted a toolkit function in a way that resulted in the function [latex]f\left(x\right)=\frac{3x+7}{x+2}[/latex]. This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.

A General Note: Rational Function

A rational function is a function that can be written as the quotient of two polynomial functions [latex]P\left(x\right) \text{and} Q\left(x\right)[/latex].

Example 3: Solving an Applied Problem Involving a Rational Function

A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?

Let t  be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:

The concentration, C , will be the ratio of pounds of sugar to gallons of water

The concentration after 12 minutes is given by evaluating [latex]C\left(t\right)[/latex] at [latex]t=\text{ }12[/latex].

This means the concentration is 17 pounds of sugar to 220 gallons of water.

At the beginning, the concentration is

Since [latex]\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05[/latex], the concentration is greater after 12 minutes than at the beginning.

Analysis of the Solution

To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:

Notice the horizontal asymptote is [latex]y=\text{ }0.1[/latex]. This means the concentration, C , the ratio of pounds of sugar to gallons of water, will approach 0.1 in the long term.

There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m.

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1.4: Inverse Functions

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• Gilbert Strang & Edwin “Jed” Herman

Learning Objectives

• Determine the conditions for when a function has an inverse.
• Use the horizontal line test to recognize when a function is one-to-one.
• Find the inverse of a given function.
• Draw the graph of an inverse function.
• Evaluate inverse trigonometric functions.

An inverse function reverses the operation done by a particular function. In other words, whatever a function does, the inverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for an inverse function to exist. We examine how to find an inverse function and study the relationship between the graph of a function and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometric functions.

Existence of an Inverse Function

We begin with an example. Given a function \(f\) and an output \(y=f(x)\), we are often interested in finding what value or values \(x\) were mapped to \(y\) by \(f\). For example, consider the function \(f(x)=x^3+4\). Since any output \(y=x^3+4\), we can solve this equation for \(x\) to find that the input is \(x=\sqrt[3]{y−4}\). This equation defines \(x\) as a function of \(y\). Denoting this function as \(f^{−1}\), and writing \(x=f^{−1}(y)=\sqrt[3]{y−4}\), we see that for any \(x\) in the domain of \(f,f^{−1}\)\(f(x))=f^{−1}(x^3+4)=x\). Thus, this new function, \(f^{−1}\), “undid” what the original function \(f\) did. A function with this property is called the inverse function of the original function.

Definition: Inverse Functions

Given a function \(f\) with domain \(D\) and range \(R\), its inverse function (if it exists) is the function \(f^{−1}\) with domain \(R\) and range \(D\) such that \(f^{−1}(y)=x\) if and only if \(f(x)=y\). In other words, for a function \(f\) and its inverse \(f^{−1}\),

\[f^{−1}(f(x))=x \nonumber \]

for all \(x\) in \(D\) and

\[f(f^{−1}(y))=y \nonumber \]

for all \(y\) in \(R\).

Note that \(f^{−1}\) is read as “\(f\) inverse.” Here, the \(−1\) is not used as an exponent so

\[f^{−1}(x)≠ \dfrac{1}{f(x)}. \nonumber \]

Figure \(\PageIndex{1}\)shows the relationship between the domain and range of \(f\) and the domain and range of \(f^{−1}\).

Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each input to exactly one output. For example, let’s try to find the inverse function for \(f(x)=x^2\). Solving the equation \(y=x^2\) for \(x\), we arrive at the equation \(x=±\sqrt{y}\). This equation does not describe \(x\) as a function of \(y\) because there are two solutions to this equation for every \(y>0\). The problem with trying to find an inverse function for \(f(x)=x^2\) is that two inputs are sent to the same output for each output \(y>0\). The function \(f(x)=x^3+4\) discussed earlier did not have this problem. For that function, each input was sent to a different output. A function that sends each input to a different output is called a one-to-one function.

Definition: One-to-One functions

We say a function \(f\) is a one-to-one function if \(f(x_1)≠f(x_2)\) when \(x_1≠x_2\).

One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the \(xy\)-plane, according to the horizontal line test , it cannot intersect the graph more than once. We note that the horizontal line test is different from the vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line test determines whether a function is one-to-one (Figure \(\PageIndex{2}\)).

Horizontal Line Test

A function \(f\) is one-to-one if and only if every horizontal line intersects the graph of \(f\) no more than once.

Example \(\PageIndex{1}\): Determining Whether a Function Is One-to-One

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

a) Since the horizontal line \(y=n\) for any integer \(n≥0\) intersects the graph more than once, this function is not one-to-one.

b) Since every horizontal line intersects the graph once (at most), this function is one-to-one.

Exercise \(\PageIndex{1}\)

Is the function \(f\) graphed in the following image one-to-one?

Use the horizontal line test.

Finding a Function’s Inverse

We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of \(f\) to elements in the range of \(f\). The inverse function maps each element from the range of \(f\) back to its corresponding element from the domain of \(f\). Therefore, to find the inverse function of a one-to-one function \(f\), given any \(y\) in the range of \(f\), we need to determine which \(x\) in the domain of \(f\) satisfies \(f(x)=y\). Since \(f\) is one-to-one, there is exactly one such value \(x\). We can find that value \(x\) by solving the equation \(f(x)=y\) for \(x\). Doing so, we are able to write \(x\) as a function of \(y\) where the domain of this function is the range of \(f\) and the range of this new function is the domain of \(f\). Consequently, this function is the inverse of \(f\), and we write \(x=f^{−1}(y)\). Since we typically use the variable \(x\) to denote the independent variable and y to denote the dependent variable, we often interchange the roles of \(x\) and \(y\), and write \(y=f^{−1}(x)\). Representing the inverse function in this way is also helpful later when we graph a function \(f\) and its inverse \(f^{−1}\) on the same axes.

Problem-Solving Strategy: Finding an Inverse Function

• Solve the equation \(y=f(x)\) for \(x\).
• Interchange the variables \(x\) and \(y\) and write \(y=f^{−1}(x)\).

Example \(\PageIndex{2}\): Finding an Inverse Function

Find the inverse for the function \(f(x)=3x−4.\) State the domain and range of the inverse function. Verify that \(f^{−1}(f(x))=x.\)

Follow the steps outlined in the strategy.

Step 1. If \(y=3x−4,\) then \(3x=y+4\) and \(x=\frac{1}{3}y+\frac{4}{3}.\)

Step 2. Rewrite as \(y=\frac{1}{3}x+\frac{4}{3}\) and let \(y=f^{−1}(x)\).Therefore, \(f^{−1}(x)=\frac{1}{3}x+\frac{4}{3}\).

Since the domain of \(f\) is \((−∞,∞)\), the range of \(f^{−1}\) is \((−∞,∞)\). Since the range of \(f\) is \((−∞,∞)\), the domain of \(f^{−1}\) is \((−∞,∞)\).

You can verify that \(f^{−1}(f(x))=x\) by writing

\(f^{−1}(f(x))=f^{−1}(3x−4)=\frac{1}{3}(3x−4)+\frac{4}{3}=x−\frac{4}{3}+\frac{4}{3}=x.\)

Note that for \(f^{−1}(x)\) to be the inverse of \(f(x)\), both \(f^{−1}(f(x))=x\) and \(f(f^{−1}(x))=x\) for all \(x\) in the domain of the inside function.

Exercise \(\PageIndex{2}\)

Find the inverse of the function \(f(x)=3x/(x−2)\). State the domain and range of the inverse function.

Use the Problem-Solving Strategy for finding inverse functions.

\(f^{−1}(x)=\dfrac{2x}{x−3}\). The domain of \(f^{−1}\) is \(\{x\,|\,x≠3\}\). The range of \(f^{−1}\) is \(\{y\,|\,y≠2\}\).

Graphing Inverse Functions

Let’s consider the relationship between the graph of a function \(f\) and the graph of its inverse. Consider the graph of \(f\) shown in Figure \(\PageIndex{3}\) and a point \((a,b)\) on the graph. Since \(b=f(a)\), then \(f^{−1}(b)=a\). Therefore, when we graph \(f^{−1}\), the point \((b,a)\) is on the graph. As a result, the graph of \(f^{−1}\) is a reflection of the graph of \(f\) about the line \(y=x\).

Example \(\PageIndex{3}\): Sketching Graphs of Inverse Functions

For the graph of \(f\) in the following image, sketch a graph of \(f^{−1}\) by sketching the line \(y=x\) and using symmetry. Identify the domain and range of \(f^{−1}\).

Reflect the graph about the line \(y=x\). The domain of \(f^{−1}\) is \([0,∞)\). The range of \(f^{−1}\) is \([−2,∞)\). By using the preceding strategy for finding inverse functions, we can verify that the inverse function is \(f^{−1}(x)=x^2−2\), as shown in the graph.

Exercise \(\PageIndex{3}\)

Sketch the graph of \(f(x)=2x+3\) and the graph of its inverse using the symmetry property of inverse functions.

The graphs are symmetric about the line \(y=x\)

Restricting Domains

As we have seen, \(f(x)=x^2\) does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of \(f\) such that the function is one-to-one. This subset is called a restricted domain . By restricting the domain of \(f\), we can define a new function \(g\) such that the domain of \(g\) is the restricted domain of \(f\) and \(g(x)=f(x)\) for all \(x\) in the domain of \(g\). Then we can define an inverse function for \(g\) on that domain. For example, since \(f(x)=x^2\) is one-to-one on the interval \([0,∞)\), we can define a new function \(g\) such that the domain of \(g\) is \([0,∞)\) and \(g(x)=x^2\) for all \(x\) in its domain. Since \(g\) is a one-to-one function, it has an inverse function, given by the formula \(g^{−1}(x)=\sqrt{x}\). On the other hand, the function \(f(x)=x^2\) is also one-to-one on the domain \((−∞,0]\). Therefore, we could also define a new function \(h\) such that the domain of \(h\) is \((−∞,0]\) and \(h(x)=x^2\) for all \(x\) in the domain of \(h\). Then \(h\) is a one-to-one function and must also have an inverse. Its inverse is given by the formula \(h^{−1}(x)=−\sqrt{x}\) (Figure \(\PageIndex{4}\)).

Example \(\PageIndex{4}\): Restricting the Domain

Consider the function \(f(x)=(x+1)^2\).

• Sketch the graph of \(f\) and use the horizontal line test to show that \(f\) is not one-to-one.
• Show that \(f\) is one-to-one on the restricted domain \([−1,∞)\). Determine the domain and range for the inverse of \(f\) on this restricted domain and find a formula for \(f^{−1}\).

a) The graph of \(f\) is the graph of \(y=x^2\) shifted left \(1\) unit. Since there exists a horizontal line intersecting the graph more than once, \(f\) is not one-to-one.

b) On the interval \([−1,∞),\;f\) is one-to-one.

The domain and range of \(f^{−1}\) are given by the range and domain of \(f\), respectively. Therefore, the domain of \(f^{−1}\) is \([0,∞)\) and the range of \(f^{−1}\) is \([−1,∞)\). To find a formula for \(f^{−1}\), solve the equation \(y=(x+1)^2\) for \(x.\) If \(y=(x+1)^2\), then \(x=−1±\sqrt{y}\). Since we are restricting the domain to the interval where \(x≥−1\), we need \(±\sqrt{y}≥0\). Therefore, \(x=−1+\sqrt{y}\). Interchanging \(x\) and \(y\), we write \(y=−1+\sqrt{x}\) and conclude that \(f^{−1}(x)=−1+\sqrt{x}\).

Exercise \(\PageIndex{4}\)

Consider \(f(x)=1/x^2\) restricted to the domain \((−∞,0)\). Verify that \(f\) is one-to-one on this domain. Determine the domain and range of the inverse of \(f\) and find a formula for \(f^{−1}\).

The domain and range of \(f^{−1}\) is given by the range and domain of \(f\), respectively. To find \(f^{−1}\), solve \(y=1/x^2\) for \(x\).

The domain of \(f^{−1}\) is \((0,∞)\). The range of \(f^{−1}\) is \((−∞,0)\). The inverse function is given by the formula \(f^{−1}(x)=−1/\sqrt{x}\).

Inverse Trigonometric Functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function. The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval \(\left[−\frac{π}{2},\frac{π}{2}\right]\). By doing so, we define the inverse sine function on the domain \([−1,1]\) such that for any \(x\) in the interval \([−1,1]\), the inverse sine function tells us which angle \(θ\) in the interval \(\left[−\frac{π}{2},\frac{π}{2}\right]\) satisfies \(\sin θ=x\). Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions , which are functions that tell us which angle in a certain interval has a specified trigonometric value.

Definition: inverse trigonometric functions

The inverse sine function, denoted \(\sin^{−1}\) or \(\arcsin\), and the inverse cosine function, denoted \(\cos^{−1}\) or \(\arccos\), are defined on the domain \(D=\{x|−1≤x≤1\}\) as follows:

\(\sin^{−1}(x)=y\)

• if and only if \(\sin(y)=x\) and \(−\frac{π}{2}≤y≤\frac{π}{2}\);

\(\cos^{−1}(x)=y\)

• if and only if \(\cos(y)=x\) and \(0≤y≤π\).

The inverse tangent function, denoted \(\tan^{−1}\) or \(\arctan\), and inverse cotangent function, denoted \(\cot^{−1}\) or \(\operatorname{arccot}\), are defined on the domain \(D=\{x|−∞<x<∞\}\) as follows:

\(\tan^{−1}(x)=y\)

• if and only if \(\tan(y)=x\) and \(−\frac{π}{2}<y<\frac{π}{2}\);

\(\cot^{−1}(x)=y\)

• if and only if \(\cot(y)=x\) and \(0<y<π\).

The inverse cosecant function, denoted \(\csc^{−1}\) or \(\operatorname{arccsc}\), and inverse secant function, denoted \(\sec^{−1}\) or \(\operatorname{arcsec}\), are defined on the domain \(D=\{x\,|\,|x|≥1\}\) as follows:

\(\csc^{−1}(x)=y\)

• if and only if \(\csc(y)=x\) and \(−\frac{π}{2}≤y≤\frac{π}{2}, \, y≠0\);

\(\sec^{−1}(x)=y\)

• if and only if \(\sec(y)=x\) and\(0≤y≤π, \, y≠π/2\).

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line \(y=x\) (Figure \(\PageIndex{5}\)).

When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate \(\cos^{−1}\left(\frac{1}{2}\right)\), we need to find an angle \(θ\) such that \(\cos θ=\frac{1}{2}\). Clearly, many angles have this property. However, given the definition of \(\cos^{−1}\), we need the angle \(θ\) that not only solves this equation, but also lies in the interval \([0,π]\). We conclude that \(\cos^{−1}\left(\frac{1}{2}\right)=\frac{π}{3}\).

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions \(\sin\left(\sin^{−1}\left(\frac{\sqrt{2}}{2}\right)\right)\) and \(\sin^{−1}(\sin(π)).\)

For the first one, we simplify as follows:

\[\sin\left(\sin^{−1}\left(\frac{\sqrt{2}}{2}\right)\right)=\sin\left(\frac{π}{4}\right)=\frac{\sqrt{2}}{2}.\nonumber \]

For the second one, we have

\[\sin^{−1}(\sin(π))=\sin^{−1}(0)=0.\nonumber \]

The inverse function is supposed to “undo” the original function, so why isn’t \(\sin^{−1}(\sin(π))=π?\) Recalling our definition of inverse functions, a function \(f\) and its inverse \(f^{−1}\) satisfy the conditions \(f(f^{−1}(y))=y\) for all \(y\) in the domain of \(f^{−1}\) and \(f^{−1}(f(x))=x\) for all \(x\) in the domain of \(f\), so what happened here? The issue is that the inverse sine function, \(\sin^{−1}\), is the inverse of the restricted sine function defined on the domain \(\left[−\frac{π}{2},\frac{π}{2}\right]\). Therefore, for \(x\) in the interval \([−\frac{π}{2},\frac{π}{2}]\), it is true that \(\sin^{−1}(\sin x)=x\). However, for values of \(x\) outside this interval, the equation does not hold, even though \(\sin^{−1}(\sin x)\) is defined for all real numbers \(x\).

What about \(\sin(\sin^{−1}y)?\) Does that have a similar issue? The answer is no. Since the domain of \(\sin^{−1}\) is the interval \([−1,1]\), we conclude that \(\sin\left(\sin^{−1}y\right)=y\) if \(−1≤y≤1\) and the expression is not defined for other values of \(y\). To summarize,

\(\sin(\sin^{−1}y)=y\) if \(−1≤y≤1\)

\(\sin^{−1}(\sin x)=x\) if \(−\frac{π}{2}≤x≤\frac{π}{2}.\)

Similarly, for the cosine function,

\(\cos(\cos^{−1}y)=y\) if \(−1≤y≤1\)

\(\cos^{−1}(\cos x)=x\) if \(0≤x≤π.\)

Similar properties hold for the other trigonometric f unctions and their inverses.

Example \(\PageIndex{5}\): Evaluating Expressions Involving Inverse Trigonometric Functions

Evaluate each of the following expressions.

• \(\sin^{−1}\left(−\frac{\sqrt{3}}{2}\right)\)
• \(\tan\left(\tan^{−1}\left(−\frac{1}{\sqrt{3}}\right)\right)\)
• \(\cos^{−1}\left(\cos\left(\frac{5π}{4}\right)\right)\)
• \(\sin^{−1}\left(\cos\left(\frac{2π}{3}\right)\right)\)
• Evaluating \(\sin^{−1}(−\sqrt{3}/2)\) is equivalent to finding the angle \(θ\) such that \(\sin θ=−\sqrt{3}/2\) and \(−π/2≤θ≤π/2\). The angle \(θ=−π/3\) satisfies these two conditions. Therefore, \(\sin^{−1}(−\sqrt{3}/2)=−π/3\).
• First we use the fact that \(\tan^{−1}(−1/\sqrt{3})=−π/6.\) Then \(\tan(-π/6)=−1/\sqrt{3}\). Therefore, \(\tan(\tan^{−1}(−1/\sqrt{3}))=−1/\sqrt{3}\).
• To evaluate \(\cos^{−1}(\cos(5π/4))\),first use the fact that \(\cos(5π/4)=−\sqrt{2}/2\). Then we need to find the angle \(θ\) such that \(\cos(θ)=−\sqrt{2}/2\) and \(0≤θ≤π\). Since \(3π/4\) satisfies both these conditions, we have \(\cos^{-1}(\cos(5π/4))=\cos^{−1}(−\sqrt{2}/2))=3π/4\).
• Since \(\cos(2π/3)=−1/2\), we need to evaluate \(\sin^{−1}(−1/2)\). That is, we need to find the angle \(θ\) such that \(\sin(θ)=−1/2\) and \(−π/2≤θ≤π/2\). Since \(−π/6\) satisfies both these conditions, we can conclude that \(\sin^{−1}(\cos(2π/3))=\sin^{−1}(−1/2)=−π/6.\)

The Maximum Value of a Function

In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.

This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable \(x\).

1. Consider the graph in Figure \(\PageIndex{6}\) of the function \(y=\sin x+\cos x.\) Describe its overall shape. Is it periodic? How do you know?

Using a graphing calculator or other graphing device, estimate the \(x\)- and \(y\)-values of the maximum point for the graph (the first such point where \(x > 0\)). It may be helpful to express the \(x\)-value as a multiple of \(π.\)

2. Now consider other graphs of the form \(y=A\sin x+B\cos x\) for various values of \(A\) and \(B.\) Sketch the graph when \(A = 2\) and \(B = 1,\) and find the \(x\)- and \(y\)-values for the maximum point. (Remember to express the \(x\)-value as a multiple of \(π\), if possible.) Has it moved?

3. Repeat for \(A = 1, \,B = 2.\) Is there any relationship to what you found in part (2)?

4. Complete the following table, adding a few choices of your own for \(A\) and \(B:\)

5. Try to figure out the formula for the \(y\)-values.

6. The formula for the \(x\)-values is a little harder. The most helpful points from the table are \((1,1),\, (1,\sqrt{3}),\, (\sqrt{3},1).\) (Hint: Consider inverse trigonometric functions.)

7. If you found formulas for parts (5) and (6), show that they work together. That is, substitute the \(x\)-value formula you found into \(y=A\sin x+B\cos x\) and simplify it to arrive at the \(y\)-value formula you found.

Key Concepts

• For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.
• If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.
• For a function \(f\) and its inverse \(f^{−1},\, f(f^{−1}(x))=x\) for all \(x\) in the domain of \(f^{−1}\) and \(f^{−1}(f(x))=x\) for all \(x\) in the domain of \(f\).
• Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.
• The graph of a function \(f\) and its inverse \(f^{−1}\) are symmetric about the line \(y=x.\)

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Number line.

• y=\frac{x^2+x+1}{x}
• f(x)=\ln (x-5)
• f(x)=\frac{1}{x^2}
• y=\frac{x}{x^2-6x+8}
• f(x)=\sqrt{x+3}
• f(x)=\cos(2x+5)
• f(x)=\sin(3x)

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• Functions A function basically relates an input to an output, there’s an input, a relationship and an output. For every input...

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Linear Functions Problems with Solutions

Linear functions are highly used throughout mathematics and are therefore important to understand. A set of problems involving linear functions , along with detailed solutions, are presented. The problems are designed with emphasis on the meaning of the slope and the y intercept.

Problem 1: f is a linear function. Values of x and f(x) are given in the table below; complete the table.

Problem 2: A family of linear functions is given by

Solution to Problem 2: a)

Problem 3: A high school had 1200 students enrolled in 2003 and 1500 students in 2006. If the student population P ; grows as a linear function of time t, where t is the number of years after 2003. a) How many students will be enrolled in the school in 2010? b) Find a linear function that relates the student population to the time t. Solution to Problem 3: a) The given information may be written as ordered pairs (t , P). The year 2003 correspond to t = 0 and the year 2006 corresponds to t = 3, hence the 2 ordered pairs (0, 1200) and (3, 1500) Since the population grows linearly with the time t, we use the two ordered pairs to find the slope m of the graph of P as follows m = (1500 - 1200) / (6 - 3) = 100 students / year The slope m = 100 means that the students population grows by 100 students every year. From 2003 to 2010 there are 7 years and the students population in 2010 will be P(2010) = P(2003) + 7 * 100 = 1200 + 700 = 1900 students. b) We know the slope and two points, we may use the point slope form to find an equation for the population P as a function of t as follows P - P1 = m (t - t1) P - 1200 = 100 (t - 0) P = 100 t + 1200

Problem 4: The graph shown below is that of the linear function that relates the value V (in $) of a car to its age t, where t is the number of years after 2000.

Problem 5: The cost of producing x tools by a company is given by

Problem 6: A 500-liter tank full of oil is being drained at the constant rate of 20 liters par minute. a) Write a linear function V for the number of liters in the tank after t minutes (assuming that the drainage started at t = 0). b) Find the V and the t intercepts and interpret them. e) How many liters are in the tank after 11 minutes and 45 seconds? Solution to Problem 6: After each minute the amount of oil in the tank deceases by 20 liters. After t minutes, the amount of oil in the tank decreases by 20*t liters. Hence if at the start there 500 liters, after t minute the amount V of oil left in the tank is given by V = 500 - 20 t b) To find the V intercept, set t = 0 in the equation V = 500 - 20 t. V = 500 liters : it is the amount of oil at the start of the drainage. To find the t intercept, set V = 0 in the equation V = 500 - 20 t and solve for t. 0 = 500 - 20 t t = 500 / 20 = 25 minutes : it is the total time it takes to drain the 500 liters of oil. c) Convert 11 minutes 45 seconds in decimal form. t = 11 minutes 45 seconds = 11.75 minutes Calculate V at t = 11.75 minutes. V(11.75) = 500 - 20*11.75 = 265 liters are in the tank after 11 minutes 45 seconds of drainage.

Problem 7: A 50-meter by 70-meter rectangular garden is surrounded by a walkway of constant width x meters.

Problem 8: A driver starts a journey with 25 gallons in the tank of his car. The car burns 5 gallons for every 100 miles. Assuming that the amount of gasoline in the tank decreases linearly, a) write a linear function that relates the number of gallons G left in the tank after a journey of x miles. b) What is the value and meaning of the slope of the graph of G? c) What is the value and meaning of the x intercept? Solution to Problem 8: a) If 5 gallons are burnt for 100 miles then (5 / 100) gallons are burnt for 1 mile. Hence for x miles, x * (5 / 100) gallons are burnt. G is then equal to the initial amount of gasoline decreased by the amount gasoline burnt by the car. Hence G = 25 - (5 / 100) x b) The slope of G is equal to 5 / 1000 and it represent the amount of gasoline burnt for a distance of 1 mile. c) To find the x intercept, we set G = 0 and solve for x. 25 - (5 / 100) x = 0 x = 500 miles : it is the distance x for which all 25 gallons of gasoline will be burnt.

Problem 9: A rectangular wire frame has one of its dimensions moving at the rate of 0.5 cm / second. Its width is constant and equal to 4 cm. If at t = 0 the length of the rectangle is 10 cm,