GMAT Quiz: Functions

6 questions · exam conditions

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FunctionsQuestion 1 of 6

If $f(x) = 2x + 1$ and $g(x) = x^2 - 3$ , what is the domain of the composite function $\frac{f(x)}{g(x)}$ ?

All real numbers except

x = \sqrt{3}

All real numbers except

x = \pm\sqrt{3}

All real numbers except

x = 3

All real numbers

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GMAT Quiz

GMAT Quiz: Functions

Practice Functions in GMAT with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If $f(x) = 2x + 1$ and $g(x) = x^2 - 3$ , what is the domain of the composite function $\frac{f(x)}{g(x)}$ ?

All real numbers except $x = \sqrt{3}$
All real numbers except $x = \pm\sqrt{3}$ (correct answer)
All real numbers except $x = 3$
All real numbers

Explanation: The function f(x)/g(x) = (2x + 1)/(x² - 3) is undefined when the denominator equals zero. Set g(x) = 0: x² - 3 = 0, so x² = 3, giving x = ±√3. Therefore the domain is all real numbers except x = √3 and x = -√3. Choice A omits the negative solution. Choice C incorrectly solves x² - 3 = 0 as x² = 3 leading to x = 3. Choice D ignores the restriction from the denominator.

Question 2

For the function $h(x) = \frac{2x - 1}{x + 3}$ , what is the horizontal asymptote?

$y = \frac{1}{3}$
$y = -3$
$y = 2$ (correct answer)
$y = 0$

Explanation: When you encounter a rational function like this, you're looking at how the function behaves as x approaches positive or negative infinity. To find the horizontal asymptote of

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

, you need to compare the degrees of the numerator and denominator polynomials. Both the numerator (2x - 1) and denominator (x + 3) are degree 1 polynomials. When the degrees are equal, the horizontal asymptote equals the ratio of the leading coefficients. The leading coefficient of the numerator is 2, and the leading coefficient of the denominator is 1, giving us

y = \frac{2}{1} = 2

. You can verify this by dividing:

\frac{2x - 1}{x + 3} = 2 - \frac{7}{x + 3}

. As x approaches infinity, the fraction

\frac{7}{x + 3}

approaches 0, leaving y approaching 2. Choice A (

y = \frac{1}{3}

) incorrectly uses the constant terms (-1 and 3) instead of the leading coefficients. Choice B (

y = -3

) mistakenly identifies the vertical asymptote (where the denominator equals zero) as the horizontal asymptote. Choice D (

y = 0

) would only be correct if the numerator's degree were less than the denominator's degree. Remember: for rational functions where numerator and denominator have equal degrees, the horizontal asymptote always equals the ratio of the leading coefficients. This pattern appears frequently on the GMAT, so focus on identifying degrees and leading coefficients quickly.

Question 3

If $f(x) = 2x^2 - 3x + 1$ and $g(x) = x - 2$ , what is the value of $f(g(3))$ ?

$0$ (correct answer)
$2$
$6$
$10$

Explanation: First, find g(3): g(3) = 3 - 2 = 1. Then find f(g(3)) = f(1): f(1) = 2(1)² - 3(1) + 1 = 2 - 3 + 1 = 0. Choice B results from calculating f(3) instead of f(g(3)). Choice C comes from incorrectly computing g(3) = 5, then f(5) = 6. Choice D results from computing f(3) + g(3) = 2 + 8 = 10.

Question 4

If $f(x) = \sqrt{x + 5}$ and the domain of $f$ is restricted so that $f$ has an inverse function, what is $f^{-1}(3)$ ?

$14$
$2$
$\sqrt{8}$
$4$ (correct answer)

Explanation: When you encounter inverse function problems, remember that finding

f^{-1}(3)

means finding the input value that produces an output of 3 when plugged into the original function

f(x)

. To find

f^{-1}(3)

, you need to solve the equation

f(x) = 3

. Setting up the equation:

\sqrt{x + 5} = 3

. Square both sides to eliminate the square root:

x + 5 = 9

. Solving for

x

x = 4

. You can verify this by checking that

f(4) = \sqrt{4 + 5} = \sqrt{9} = 3

✓ Now let's examine why the other answers are incorrect: Choice A (

14

) likely comes from incorrectly setting up

\sqrt{x + 5} = 3

, then solving

x + 5 = 3 + 5 = 8

, leading to

x = 14 - 5 = 9

. This reflects confusion about inverse operations. Choice B (

2

) might result from the error

x + 5 = 3

, giving

x = -2

, then taking the absolute value to get

2

. This shows misunderstanding of how to handle the square root equation. Choice C (

\sqrt{8}

) could come from trying to work backwards incorrectly, perhaps thinking

f^{-1}(3) = \sqrt{3 + 5} = \sqrt{8}

. This represents applying the original function instead of finding its inverse. The correct answer is D. Strategy tip: For inverse function problems, always set

f(x)

equal to the given output value and solve for

x

. Don't try to manipulate the function formula itself—work with the equation

f(x) = \text{target value}

Question 5

The function $p(x) = ax^2 + bx + c$ passes through the points $(1, 4)$ , $(2, 9)$ , and $(3, 16)$ . What is the value of $a + b + c$ ?

$1$
$4$ (correct answer)
$9$
$16$

Explanation: Notice that a + b + c = p(1) = 4, since when x = 1, p(x) = a(1)² + b(1) + c = a + b + c. We can verify this by setting up the system: p(1) = a + b + c = 4, p(2) = 4a + 2b + c = 9, p(3) = 9a + 3b + c = 16. From the first equation, we immediately get a + b + c = 4. Choice A incorrectly assumes the function is p(x) = x². Choice C results from confusing a + b + c with p(2). Choice D results from confusing a + b + c with p(3).

Question 6

The function $k(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 2x - 1 & \text{if } x > 1 \end{cases}$ is defined piecewise. What is $k(k(0.5))$ ?

$\frac{1}{4}$
$0$
$\frac{1}{16}$ (correct answer)
$\frac{1}{2}$

Explanation: Piecewise functions require you to evaluate which piece of the function applies based on the input value. When dealing with composite functions like

k(k(0.5))

, you must work from the inside out. First, evaluate

k(0.5)

. Since

0.5 \leq 1

, you use the first piece:

k(0.5) = (0.5)^2 = 0.25 = \frac{1}{4}

. Now evaluate

k(0.25)

. Since

0.25 \leq 1

, you again use the first piece:

k(0.25) = (0.25)^2 = 0.0625 = \frac{1}{16}

. Therefore,

k(k(0.5)) = \frac{1}{16}

. Looking at the wrong answers: Choice A gives

\frac{1}{4}

, which is the value of

k(0.5)

but not the final answer—this represents stopping after the first step. Choice B gives

0

, which doesn't result from any reasonable calculation with this function. Choice D gives

\frac{1}{2}

, which might come from incorrectly applying the second piece of the function or making an arithmetic error. The key strategy with piecewise functions is to carefully check which condition your input satisfies before applying the corresponding rule. For composite functions, work step by step from the innermost function outward, and don't forget to re-check which piece applies at each step. The intermediate result from the inner function becomes the input for the outer function evaluation.