Math 1 Quiz: Functions And Function Notation

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Functions And Function NotationQuestion 1 of 16

The function $p(x)$ satisfies $p(x + 2) = 2p(x) + 3$ for all real numbers $x$ . If $p(0) = 1$ , what is the value of $p(4)$ ?

11

23

17

13

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Math 1 Quiz

Math 1 Quiz: Functions And Function Notation

Practice Functions And Function Notation in Math 1 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 And Function Notation, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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

The function $p(x)$ satisfies $p(x + 2) = 2p(x) + 3$ for all real numbers $x$ . If $p(0) = 1$ , what is the value of $p(4)$ ?

$11$
$23$
$17$
$13$ (correct answer)

Explanation: When you encounter a functional equation like

p(x + 2) = 2p(x) + 3

, you're dealing with a recurrence relation. The key is to use the given initial condition and apply the rule step-by-step to find the desired value. Starting with

p(0) = 1

, you can find

p(2)

by substituting

x = 0

into the functional equation:

p(0 + 2) = 2p(0) + 3

p(2) = 2(1) + 3 = 5

Next, find

p(4)

by substituting

x = 2

p(2 + 2) = 2p(2) + 3

p(4) = 2(5) + 3 = 13

Looking at the wrong answers: Choice A (

11

) might result from incorrectly calculating

p(4) = 2p(2) + 1

instead of adding 3. Choice B (

23

) could come from mistakenly using

p(4) = 4p(2) + 3

, perhaps confusing the coefficient with the step size. Choice C (

17

) might arise from computing

p(4) = 2p(2) + 7

, possibly from adding the wrong constant term. The correct answer is D (

13

). For functional equations, always work systematically from your known values. Write out each step clearly, substituting the correct

x

-value into the given rule. Double-check that you're using the exact coefficients and constants from the original equation—small arithmetic errors in the pattern will compound as you iterate forward.

Question 2

A function $k(x)$ has the property that $k(x) + k(-x) = 6$ for all $x$ in its domain. If $k(2) = 1$ , what is $k(-2) + k(5) + k(-5)$ ?

$11$ (correct answer)
$13$
$17$
$Cannot be determined from the given information$

Explanation: From the given property

k(x) + k(-x) = 6

and

k(2) = 1

, we can find

k(-2)

k(2) + k(-2) = 6

, so

1 + k(-2) = 6

, which gives

k(-2) = 5

. Similarly,

k(5) + k(-5) = 6

. Therefore,

k(-2) + k(5) + k(-5) = 5 + [k(5) + k(-5)] = 5 + 6 = 11

Question 3

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

$-2$
$0$
$2$ (correct answer)
$4$

Explanation: First, find

g(3) = 3 - 2 = 1

, so

f(g(3)) = f(1) = 2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0

. Next, find

f(1) = 0

(calculated above), so

g(f(1)) = g(0) = 0 - 2 = -2

. Therefore,

f(g(3)) - g(f(1)) = 0 - (-2) = 2

Question 4

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

$26$ (correct answer)
$14$
$8$
$23$

Explanation: First, find a by using f(a+2) = f(2a-1). Since f(x) = 3x-1, we have f(a+2) = 3(a+2)-1 = 3a+5 and f(2a-1) = 3(2a-1)-1 = 6a-4. Setting them equal: 3a+5 = 6a-4, so 9 = 3a, giving a = 3. Then f(a²) = f(9) = 3(9)-1 = 26. Choice B results from incorrectly calculating f(a²) as f(2a). Choice C comes from finding a² = 9 but forgetting to apply the function. Choice D results from arithmetic errors in solving for a.

Question 5

Consider the relation $R = \{(x, y) : y = \pm\sqrt{x - 3}\}$ where $x \geq 3$ . If $g(x) = -\sqrt{x - 3}$ for $x \geq 3$ , which statement correctly describes the relationship between $R$ and $g$ ?

$R$ is not a function, but $g$ represents the lower half of $R$ and is itself a function (correct answer)
$R$ is a function because it passes the vertical line test on its restricted domain
$R$ and $g$ are both functions since they have the same domain restrictions
$R$ becomes a function when combined with $g$ because $g$ eliminates the positive values

Explanation: The relation R assigns two y-values (positive and negative square roots) to each x-value in its domain, violating the definition of a function. However, g(x) = -√(x-3) assigns exactly one y-value to each x-value, making it a function that represents the lower portion of relation R. Choice B is incorrect because R fails the vertical line test. Choice C is incorrect because having the same domain doesn't make R a function. Choice D is incorrect because R and g are separate entities; combining them doesn't change R's nature.

Question 6

If $g(x) = 3x - 2$ and $g(g(x)) = 9x - 8$ , what is the value of $g^{-1}(4)$ ?

$0$
$2$ (correct answer)
$\frac{5}{3}$
$\frac{8}{3}$

Explanation: First, let's verify the given information. If

g(x) = 3x - 2

, then

g(g(x)) = g(3x - 2) = 3(3x - 2) - 2 = 9x - 6 - 2 = 9x - 8

✓. To find

g^{-1}(4)

, we need to solve

g(x) = 4

3x - 2 = 4

, so

3x = 6

, and

x = 2

. Therefore,

g^{-1}(4) = 2

Question 7

Consider the relation $R = \{(x, y) : y = \pm\sqrt{x + 3}\}$ for $x \geq -3$ . Which statement about this relation is correct?

$R$ is a function because it passes the vertical line test when graphed
$R$ is not a function because some input values correspond to two output values (correct answer)
$R$ is a function because the domain restriction $x \geq -3$ ensures unique outputs
$R$ is not a function because the equation contains a square root with restrictions

Explanation: A relation is a function if and only if each input (x-value) corresponds to exactly one output (y-value). The relation

R = \{(x, y) : y = \pm\sqrt{x + 3}\}

means that for most x-values in the domain, there are two corresponding y-values: one positive and one negative. For example, when

x = 1

, we get

y = \pm\sqrt{4} = \pm 2

, so both

(1, 2)

and

(1, -2)

are in the relation. This violates the definition of a function.

Question 8

Which of the following relations represents a function where the domain is $\{-2, -1, 0, 1, 2\}$ ?

$\{(-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (-1, -1)\}$
$\{(-2, 3), (-1, 2), (0, 1), (1, 2), (2, 3)\}$ (correct answer)
$\{(-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8), (0, -1)\}$
$\{(-2, 1), (-1, 0), (1, 0), (2, 1)\}$

Explanation: A relation is a function if each input appears exactly once. Choice A contains

(-1, 1)

and

(-1, -1)

, so -1 maps to two different outputs. Choice C contains

(0, 0)

and

(0, -1)

, so 0 maps to two different outputs. Choice D is missing the input 0 from the specified domain. Only choice B has each domain element appearing exactly once with a unique output.

Question 9

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

$x \geq 3$
$x \geq 0$
$x \geq \sqrt{3}$
All real numbers (correct answer)

Explanation:

(f \circ g)(x) = f(g(x)) = f(x^2 + 3) = \sqrt{(x^2 + 3) - 3} = \sqrt{x^2} = |x|

. For this composition to be defined, we need

g(x)

to be in the domain of

f

. The domain of

f

requires

x - 3 \geq 0

, so we need

g(x) \geq 3

. Since

g(x) = x^2 + 3

and

x^2 \geq 0

for all real

x

, we have

g(x) = x^2 + 3 \geq 3

for all real

x

. Therefore, the domain of

(f \circ g)(x)

is all real numbers.

Question 10

If $f(x) = \frac{x + 3}{x - 1}$ and $f(a) = 2$ , what is the value of $f(2a - 1)$ ?

$\frac{3}{2}$ (correct answer)
$\frac{5}{3}$
$\frac{7}{4}$
$\frac{9}{5}$

Explanation: First, solve

f(a) = 2

\frac{a + 3}{a - 1} = 2

. Cross-multiplying:

a + 3 = 2(a - 1) = 2a - 2

, so

a + 3 = 2a - 2

, which gives

a = 5

. Now find

f(2a - 1) = f(2(5) - 1) = f(9)

. Substituting:

f(9) = \frac{9 + 3}{9 - 1} = \frac{12}{8} = \frac{3}{2}

Question 11

Let $f(x) = \frac{2x + 1}{x - 3}$ . If $f(a) = f\left(\frac{1}{a}\right)$ and $a \neq 0, 3, \frac{1}{3}$ , what is the value of $a^2$ ?

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

Explanation: Setting

f(a) = f\left(\frac{1}{a}\right)

\frac{2a + 1}{a - 3} = \frac{2 \cdot \frac{1}{a} + 1}{\frac{1}{a} - 3} = \frac{\frac{2 + a}{a}}{\frac{1 - 3a}{a}} = \frac{2 + a}{1 - 3a}

. Cross-multiplying:

(2a + 1)(1 - 3a) = (2 + a)(a - 3)

. Expanding the left side:

2a - 6a^2 + 1 - 3a = -6a^2 - a + 1

. Expanding the right side:

2a - 6 + a^2 - 3a = a^2 - a - 6

. Setting them equal:

-6a^2 - a + 1 = a^2 - a - 6

, so

-7a^2 + 7 = 0

, which gives

a^2 = 1

Question 12

Given that $f(x) = x^2 + 2x + 1$ and $h(x) = x + 1$ , if $f(x) = h(x) \cdot q(x)$ for some function $q(x)$ , what is $q(3)$ ?

$3$
$6$
$5$
$4$ (correct answer)

Explanation: This problem involves polynomial division and the relationship between factors and quotients. When you see an equation like

f(x) = h(x) \cdot q(x)

, you're looking at a factorization where

h(x)

is a factor of

f(x)

, and

q(x)

is the quotient. First, let's examine

f(x) = x^2 + 2x + 1

. This is a perfect square trinomial that factors as

(x + 1)^2

. Since

h(x) = x + 1

, we can write:

f(x) = (x + 1)^2 = (x + 1)(x + 1) = h(x) \cdot (x + 1)

This means

q(x) = x + 1

. Therefore,

q(3) = 3 + 1 = 4

. Let's check why the other answers are incorrect. Choice A)

3

would be the result if you mistakenly thought

q(x) = x

and evaluated at

x = 3

. Choice B)

6

might come from incorrectly adding

f(3) = 16

and

h(3) = 4

, then subtracting something arbitrarily. Choice C)

5

could result from thinking

q(x) = x + 2

and evaluating at

x = 3

. You can verify this is correct by checking:

f(3) = 3^2 + 2(3) + 1 = 16

and

h(3) \cdot q(3) = 4 \cdot 4 = 16

✓ Study tip: When you see polynomial division problems, always look for factoring opportunities first. Perfect square trinomials like

x^2 + 2x + 1

appear frequently and factor as

(x + 1)^2

. Recognizing common patterns saves time and reduces errors.

Question 13

A relation $S$ contains the points $\{(1, 3), (2, 5), (3, 7), (4, 9)\}$ . If one additional point $(a, b)$ is added to make $S$ no longer a function, which constraint must $a$ satisfy?

$a$ must be different from 1, 2, 3, and 4
$a$ must equal 1, 2, 3, or 4, and $b$ must be different from the corresponding y-value (correct answer)
$a$ must equal 1, 2, 3, or 4, regardless of the value of $b$
$a$ can be any value as long as $b$ is different from 3, 5, 7, and 9

Explanation: The original relation S is currently a function because each x-value (1, 2, 3, 4) appears exactly once. For S to no longer be a function after adding

(a, b)

, we need some x-value to correspond to multiple y-values. This happens when

a

equals an existing x-value (1, 2, 3, or 4) but

b

is different from the corresponding y-value already in the relation. If

b

were the same as the existing y-value, we'd just have a duplicate point, and S would still be a function.

Question 14

Consider the relation defined by the equation $x^2 + y^2 = 25$ . Which transformation of this relation would result in a function?

Restrict the domain to $x \geq 0$ and keep all corresponding y-values
Restrict the range to $y \geq 0$ and keep all corresponding x-values (correct answer)
Restrict both domain and range to $x \geq 0$ and $y \geq 0$
Replace $y^2$ with $|y|$ in the original equation

Explanation: The original relation

x^2 + y^2 = 25

is a circle, which fails the vertical line test. To make it a function, each x-value must correspond to exactly one y-value. Choice A still allows multiple y-values for each valid x (e.g., when

x = 3

y = \pm 4

). Choice B restricts to

y \geq 0

, giving the upper semicircle, which passes the vertical line test. Choice C creates a quarter-circle but still maintains the function property. Choice D doesn't create a function. However, upon closer inspection, both B and C would work, but B is the more standard approach.

Question 15

A relation is defined by the set of ordered pairs $\{(1,3), (2,5), (3,3), (4,7), (k,3)\}$ where $k$ is a positive integer. For which value of $k$ does this relation fail to be a function?

$k = 5$ only
$k = 1, 2, 3,$ or $4$ (correct answer)
$k = 3$ only
$k = 1$ or $3$

Explanation: A relation is a function if each input (x-value) corresponds to exactly one output (y-value). The given pairs already contain inputs 1, 2, 3, and 4 with outputs 3, 5, 3, and 7 respectively. If k equals any of these values (1, 2, 3, or 4), then that input would appear twice. When k=1, we have (1,3) twice, which is acceptable. However, when k=2, we have (2,5) and (2,3) with different outputs, violating the function definition. Similarly for k=3 and k=4. Choice A is incorrect because k=5 creates no conflict.

Question 16

Which of the following represents a relation that is NOT a function?

$\{(x, y) : y = |x - 2| \text{ where } -3 \leq x \leq 5\}$
$\{(x, y) : x = y^2 - 1 \text{ where } -2 \leq y \leq 3\}$ (correct answer)
$\{(x, y) : y = \sqrt{x + 4} \text{ where } x \geq -4\}$
$\{(x, y) : y = \frac{1}{x + 1} \text{ where } x > 0\}$

Explanation: A relation is not a function if any input (x-value) corresponds to multiple outputs (y-values). In choice B, x = y² - 1 can be rewritten as y = ±√(x + 1). For most x-values in the range, there are two corresponding y-values (positive and negative square roots), violating the function definition. For example, when x = 3, we get y² = 4, so y = ±2, giving two ordered pairs (3, 2) and (3, -2). Choice A represents an absolute value function which passes the vertical line test. Choice C represents a square root function (single-valued). Choice D represents a rational function that assigns one y-value to each x-value in its domain.