Evaluating Functions | SAT Advanced Math | SAT Math

Introduction

Understanding how to evaluate a function requires an understanding of what a function is and how its parts relate, so we will address those concepts in this lesson. To evaluate a function is, traditionally defined, to substitute a number, variable, or expression for $x$ in the function and simplify the result. Because the SAT has these kinds of problems but also very similar ones in which the substituted value replaces $f (x)$ instead of $x$ , we will extend the concept to cover that idea as well.

Functions on the SAT may be linear, quadratic, or exponential, as well as (occasionally) cubic (highest power = $3$ ). For this reason, this lesson overlaps significantly with the lessons on linear, quadratic, and exponential equations.

Approach Question

The function $f$ is defined as $f (x) = (a - x) (x - 11)$ , where $a$ is a constant. In the $x y$ -plane, the graph of $y = f (x)$ passes through the point $(19, 0)$ . What is the value of $f (0)$ ? (Note: this is a free-response question.)

Explanation

Before addressing the Approach question in particular, we need to begin by describing functions in general. As you have likely learned in school, a function is typically defined as a relation of a set of inputs to possible outputs, with only one output possible for each input. To understand this abstract definition, it helps to define “input” and “output”. Think of a vending machine with different prices assigned to different items. If you put amount A into the machine, you get a bag of chips. If you put amount B in, you get a candy bar. If you put amount C in, you get a soda. Different inputs yield different outputs.

A function works in a similar way, except that the $x$ variable (known as the independent variable in a graph) is assigned as the input and the $y$ variable (the dependent variable on a graph) is assigned as the output. Further, the language $f (x)$ is often used instead of $y$ ; this stands for “function of x,” which is another way of saying “the result when x is inputted”. So you can think of the different inputs in the vending machine scenario as different possible values of $x$ , while the $y$ (or $f (x)$ ) part of the equation represents the “results” you receive: chips, candy bar, soda. Finally, even though $f$ stands for function, other letters are often used to stand in the place of $f$ , most commonly $g$ and $h$ (so $g (x)$ and $h (x)$ ). As long as the format has one letter standing before another letter in parentheses, you should assume that’s function notation. It’s not two different variables - a frequent point of confusion for students learning about functions!

With functions defined, we now turn to what it means to evaluate a function. The “evaluating” part of the Approach question occurs when it asks for the value of $f (0)$ . Since the $0$ is inside the parentheses where the $x$ in $f (x)$ would be, the $0$ must be the input of this function we are supposed to plug in. If the “parent function” (the original function giving the “rules of the game”) has multiple instances of $x$ , then $0$ must be substituted for all such instances.

Since our function here is $f (x) = (a - x) (x - 11)$ , we can plug in $0$ for all of the $x$ ’s we see. The result is $f (0) = a (- 11)$ . Remember that $f (0)$ just means “the result when $0$ is substituted for $x$ ”; $f (0)$ is simply reminding us what we are doing at this stage of evaluating the function. If we rewrite $a (- 11)$ as $- 11 a$ , are we finished? The answer is no, because we have an unknown $a$ left, and unknowns cannot be entered as student-produced responses on the SAT. How do we find what a is? Unfortunately, we have to backtrack a bit, which we’re doing in this case to show you how one approach to the problem is more efficient than the other.

Instead of plugging in $0$ immediately for $x$ , it makes more sense to plug in the point $(19, 0)$ , because that provides a numeric value for both input and output. If we plug in $0$ for $f (x)$ and $19$ for all instances of $x$ , the result is $0 = (a - 19) (8)$ . Distributing the $8$ , we get $8 a - 152 = 0$ ; we can add $152$ to both sides and then divide by $8$ , revealing that $a = 19$ . (Looking back, do you see why $a$ must equal $19$ ? When we have $0 = (a - 19) (8)$ , the only way to get $0$ on the right side is for $a$ to be $19$ .

Now that we have resolved what $a$ is, we can rewrite the function with the value of a and then evaluate it at $x = 0$ . Hence:

$f (x) f (0) f (0) = (19 - x) (x - 11) = (19) (- 11) = - 201$

The answer is -201.

Definitions

Input: The number, variable, or expression inserted into a function at all locations of $x$ . Since $x$ is inside the parentheses in $f (x)$ and $x$ represents the input, we can remember the relationship with “inside=input”.
Output: The result when an input is substituted and the function is simplified. The output may be represented by either $y$ or $f (x)$ . If a problem provides you with the result (output) while leaving the input unknown, you must substitute the given number for $y$ or $f (x)$ and solve for $x$ .
Parent Function: If multiple functions are present, or even statements of specific value such as $f (0) = 10$ , it can be helpful to refer to the original function as the “parent” function that governs the action in the problem.

Topics for Cross-Reference

Variations

As noted in the introduction, the main variation in “evaluate” problems is the sort of function being evaluated at a given value. Use this lesson as an opportunity to review your work on linear equations and anticipate your work on quadratic and exponential functions.

Strategy Insights

It’s all about input versus output. Make triple sure you which of these question has provided and which it has left unknown and asked you to find.
If a question has constants–letters other than $x$ or $y$ meant to stand for unchanging values–it typically makes sense to use the given information to algebraically determine the value of the constant. Once you have replaced the constant’s letter with its numerical value, you will typically be in good shape to evaluate the function.

Flashcard Fodder

“Inside=input”. Perhaps worth a flashcard?

Sample Questions

Difficulty 1

The function $h$ is defined as $h (x) = 93 - 17 x$ . What is the value of $h (- 3)$ ?

A. 11
B. 12
C. 121
D. 144

(spoiler)

The answer is 12. The most straightforward sort of “evaluating” question will use a format such as $h (- 3)$ . Remember that 1) functions can use other letters besides $f$ as their indicator (most commonly $g$ and $h$ ), and 2) that the number inside the parentheses is the input, which means it substitutes for $x$ .

So, substituting for x, we come up with $h (- 3) = 93 - 17 (- 3)$ , which simplifies to $93 + 51$ , then $144$ , then $12$ .

Difficulty 2

Ivy is going door to door selling cookies. She models her anticipated profit with the function $p (c) = 7 c - 150$ , where $p (c)$ is the profit, in dollars, and $c$ is the number of boxes of cookies she sells. To achieve a profit of $$410$ , how many boxes of cookies must Ivy sell? (Note: this is a free-response question.)

(spoiler)

The answer is 80. Word problems involving functions will often contain, as their primary challenge, the question of whether the value given should be plugged in for $x$ or for $f (x)$ . Is it in the input or the output–the initial value or the result? In this case, the structure of the problem makes clear that $c$ represents the input, so the number of boxes of cookies, if given, would be plugged in for $c$ . But in this case, we instead get the profit, which is represented by $p (c)$ . Remember that $p (c)$ is not two different variables but one concept, representing the result when a certain value is plugged in for $c$ . In this case, then, we take the given value of $410$ and substitute it for $p (c)$ , then solve for $c$ to get the number of boxes of cookies required. Here we go:

$41056080 = 7 c - 150 = 7 c = c$

Difficulty 3

The function $g$ is defined by $g (x) = - 6 (1 + x^{3}) - \frac{3}{2}$ . What is the value of $g (\frac{1}{2})$ ?

A. $- \frac{33}{4}$
B. $- \frac{27}{4}$
C. $- \frac{21}{4}$
D. $- \frac{15}{4}$

(spoiler)

The answer is $- \frac{33}{4}$ . This question doesn’t differ in form from a lower-difficulty question (we are given an input to substitute for $x$ in one place); what sets it apart is the challenge of manipulating fractions. First, we must carefully raise $\frac{1}{2}$ to the third power; then we must add it to $1$ using a common denominator; then we must multiply it by $- 6$ ; finally, we will need a common denominator again in order to subtract $\frac{3}{2}$ from our previous result. Finally, we’ll have to reduce our fraction to match one of the answers. Here’s the process:

$- 6 (1 + (1/2)^{3}) - 3/2 - 6 (1 + 1/8) - 3/2 - 6 (9/8) - 3/2 - 54/8 - 3/2 - 54/8 - 12/8 - 66/8 - 33/4$

Difficulty 4

The exponential function $f$ is defined by $f (x) = 13 a^{x}$ , where $a$ is a constant. If $f (2) = 637$ , what is the value of $f (3)$ ? (Note: this is a free-response question.)

(spoiler)

The answer is 4,459. In this question, the value of the constant $a$ is what unlocks the solving process. Before evaluating the function at $3$ , as the $f (3)$ asks us to do, we need to interpret the statement that $g (2) = 637$ . Following function logic, we infer that plugging in $2$ for $x$ should yield $637$ in the place of $f (x)$ . This should allow us to solve for $a$ :

$637497 = 13 a^{2} = a^{2} = a$

Now we can write the function as $f (x) = 13 (7^{x})$ . If we now plug in $3$ for $x$ and use our calculator, we will get the correct result. As a way of checking, note that $4, 459$ is simply $637 \times 7$ . This makes sense because we have raised the exponent by one, meaning we multiply by the base one additional time, and that base is $7$ .

Difficulty 5

The function $h$ is defined as $h (x) = j (x + k)^{2}$ , where $j$ and $k$ are constants. In the $x y$ -plane, the graph of $y = h (x)$ passes through the point $(- 14, 0)$ and $h (14) < 0$ . Which of the following must be true?

A. $j < k$
B. $j > k$
C. $h (0) = 14$
D. $h (0) = - 14$

(spoiler)

The answer is $j < k$ . There is an aspect of function evaluation going on here in the understanding that $h (14) < 0$ and that $h (- 14) = 0$ (which is another way of saying the graph passes through $(- 14, 0)$ ). Using this information and logic, we can reason toward what must be true.

As we’ll see in the lesson on exponential functions, the concept of intercepts becomes especially important on certain challenging function problems. We know that $x$ -intercepts sit on the $x$ -axis and take the form $(x, 0)$ , while $y$ -intercepts sit on the $y$ -axis and take the form $(0, y)$ . This recognition shows us that $(- 14, 0)$ is an $x$ -intercept here; the value of $0$ will be especially helpful.

The question is, since $0 = j (- 14 + k)^{2}$ (plugging the coordinates $(- 14, 0)$ into the function), how can we make the right side equal to zero? Since there are two quantities multiplied together, at least one of the quantities must be equal to zero. That tells us that either $j = 0$ or $- 14 - k = 0$ . If the latter is true, then $k$ must equal $- 14$ . Here’s where the knowledge that $h (- 14) < 0$ comes in handy. If we plug in $14$ for $x$ , we get $j (14 + k)^{2}$ , and we know that the value of this expression must be negative. How will it become negative? Certainly not through a squared term, which must always be nonnegative because squaring any number results in either $0$ or a positive value. The only conclusion left is that $j$ is negative; if multiplied by the inevitably positive $(14 + k)^{2}$ , a negative $j$ will ensure the whole right side is negative.

Consider now our former reasoning: that either $j = 0$ or $- 14 + k = 0$ . We have discovered that $j$ cannot be equal to $0$ ; it is negative. This means that $- 14 + k = 0$ ; adding $k$ to both sides shows us that $k = 14$ . If $j$ is negative and $k = 14$ , then $j < k$ .

For Reflection

Compared to how you’ve done on these sorts of questions in the past, how did you do with evaluating functions in this lesson? What further practice is necessary?
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
Do you understand the difference between input and output of a function, and when to plug in values for which? If not, take some time to reread the parts of the lesson that deal with those concepts.

Introduction

Approach Question

The function $f$ is defined as $f (x) = (a - x) (x - 11)$ , where $a$ is a constant. In the $x y$ -plane, the graph of $y = f (x)$ passes through the point $(19, 0)$ . What is the value of $f (0)$ ? (Note: this is a free-response question.)

Explanation

Now that we have resolved what $a$ is, we can rewrite the function with the value of a and then evaluate it at $x = 0$ . Hence:

$f (x) f (0) f (0) = (19 - x) (x - 11) = (19) (- 11) = - 201$

The answer is -201.

Definitions

Input: The number, variable, or expression inserted into a function at all locations of $x$ . Since $x$ is inside the parentheses in $f (x)$ and $x$ represents the input, we can remember the relationship with “inside=input”.
Output: The result when an input is substituted and the function is simplified. The output may be represented by either $y$ or $f (x)$ . If a problem provides you with the result (output) while leaving the input unknown, you must substitute the given number for $y$ or $f (x)$ and solve for $x$ .
Parent Function: If multiple functions are present, or even statements of specific value such as $f (0) = 10$ , it can be helpful to refer to the original function as the “parent” function that governs the action in the problem.

Topics for Cross-Reference

Variations

Strategy Insights

It’s all about input versus output. Make triple sure you which of these question has provided and which it has left unknown and asked you to find.
If a question has constants–letters other than $x$ or $y$ meant to stand for unchanging values–it typically makes sense to use the given information to algebraically determine the value of the constant. Once you have replaced the constant’s letter with its numerical value, you will typically be in good shape to evaluate the function.

Flashcard Fodder

“Inside=input”. Perhaps worth a flashcard?

Sample Questions

Difficulty 1

The function $h$ is defined as $h (x) = 93 - 17 x$ . What is the value of $h (- 3)$ ?

A. 11
B. 12
C. 121
D. 144

(spoiler)

So, substituting for x, we come up with $h (- 3) = 93 - 17 (- 3)$ , which simplifies to $93 + 51$ , then $144$ , then $12$ .

Difficulty 2

Ivy is going door to door selling cookies. She models her anticipated profit with the function $p (c) = 7 c - 150$ , where $p (c)$ is the profit, in dollars, and $c$ is the number of boxes of cookies she sells. To achieve a profit of $$410$ , how many boxes of cookies must Ivy sell? (Note: this is a free-response question.)

(spoiler)

$41056080 = 7 c - 150 = 7 c = c$

Difficulty 3

The function $g$ is defined by $g (x) = - 6 (1 + x^{3}) - \frac{3}{2}$ . What is the value of $g (\frac{1}{2})$ ?

A. $- \frac{33}{4}$
B. $- \frac{27}{4}$
C. $- \frac{21}{4}$
D. $- \frac{15}{4}$

(spoiler)

$- 6 (1 + (1/2)^{3}) - 3/2 - 6 (1 + 1/8) - 3/2 - 6 (9/8) - 3/2 - 54/8 - 3/2 - 54/8 - 12/8 - 66/8 - 33/4$

Difficulty 4

The exponential function $f$ is defined by $f (x) = 13 a^{x}$ , where $a$ is a constant. If $f (2) = 637$ , what is the value of $f (3)$ ? (Note: this is a free-response question.)

(spoiler)

$637497 = 13 a^{2} = a^{2} = a$

Difficulty 5

The function $h$ is defined as $h (x) = j (x + k)^{2}$ , where $j$ and $k$ are constants. In the $x y$ -plane, the graph of $y = h (x)$ passes through the point $(- 14, 0)$ and $h (14) < 0$ . Which of the following must be true?

A. $j < k$
B. $j > k$
C. $h (0) = 14$
D. $h (0) = - 14$

(spoiler)

For Reflection

Compared to how you’ve done on these sorts of questions in the past, how did you do with evaluating functions in this lesson? What further practice is necessary?
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
Do you understand the difference between input and output of a function, and when to plug in values for which? If not, take some time to reread the parts of the lesson that deal with those concepts.