McGraw-Hill Education GRE 2019

Going from Output to Input

In the preceding example, you were given the formula for the function and its

input, and you were asked to arrive at the output. Sometimes you will be given the

formula for the function and its output, and you will be asked to arrive at the input.

Look at the following example:

For all positive numbers, x, the function f is defined as f(x) = (x + 3)^2 – 5.

If f(a) = 59, then a =?

First, note that the function only applies to positive inputs, so you know your

answer will be positive. To solve for a, you go through a similar substitution

process as presented earlier.

Step 1: Plug a into the function and arrive at f(a) = (a + 3)^2 – 5

Step 2: Set the second expression equal to 59: (a + 3)^2 – 5 = 59

Step 3: Solve for a: (a + 3)^2 = 64 → (a + 3) = 8 → a = 5

Certain function questions will give you two functions and ask you to use the output

of one function as the input for the other function. Look at the following example:

For all numbers, x, f(x) is defined by f(x) = x^2 + 12, and g(x) is defined

by g(x) = √x + 5. What is the value of f(g4)?

This question wants you to use the output of g(4) as your input for the

function f. When working with a compound function, always go inside-out:

Step 1: Solve for g(4): g(4) = √4 + 5 = 3.

Step 2: Substitute the output of g(4) into the function f: f(3) = 3^2 + 12 = 21.

Symbolism

A different type of function question will define a formula using symbols rather

than the notation covered in the previous section. Such questions are often a source

of intimidation for students, but the process for these questions is almost identical

to the process for normal function questions.

If the operation * is defined for all numbers, a, by a* = 2a^3 , and (b + 3)* = 54,

then b =?

Step 1: Understand the formula: you should substitute whatever is in front of

the multiplication sign for a in the formula 2a^3.

Step 2: Substitute (b +3) for a: 2(b + 3)^3 = 54.

Step 3: Solve for b:

2(b + 3)^3 = 54

(b + 3)^3 = 27

b + 3 = 3

b = 0

CHAPTER 11 ■ ALGEBRA 293

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