ACT Math — Functions Domain Test

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Question 1

If \(f(x)=3x-4\), what is \(f(5)\)?

A) 7
B) 11
C) 15
D) 19

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Substitute \(x=5\): \(f(5)=3(5)-4=11\).

Answer: B

Question 2

Which relation defines \(y\) as a function of \(x\)?

F) \(\{(1,2),(1,5),(3,4)\}\)
G) \(\{(0,3),(2,3),(4,3)\}\)
H) \(\{(-1,2),(-1,4),(2,5)\}\)
J) \(\{(2,1),(2,2),(2,3)\}\)

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A relation is a function when each input has exactly one output.

Only choice G has no input paired with two different outputs.

Answer: G

Question 3

The graph of \(f\) is shown. Which statement must be true?

A) \(f(2)<0\)
B) \(f(-2)>0\)
C) \(f(2)=f(-2)\)
D) \(f(2)>f(-2)\)

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The graph shows \(f(2)=1\) and \(f(-2)=-1\): \(f(2)>f(-2)\).

Answer: D

Question 4

If \(g(x)=x^2+2\), what is \(g(-3)\)?

F) 5
G) 7
H) 9
J) 11

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Substitute \(x=-3\): \(g(-3)=(-3)^2+2=11\).

Answer: J

Question 5

The function \(h(x)=2^x\) is an example of which type of function?

A) Linear
B) Quadratic
C) Exponential
D) Piecewise

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The variable \(x\) appears in the exponent.

The function is exponential.

Answer: C

Question 6

The graph of \(f\) is shown only for the closed interval from \(x=-3\) to \(x=4\). What is the domain of \(f\)?

F) \([-3,4]\)
G) \((-3,4)\)
H) \([-2,3]\)
J) \((-2,3)\)

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The graph exists from \(x=-3\) through \(x=4\), and both endpoints are included.

The domain is \([-3,4]\).

Answer: F

Question 7

If \(f(x)=x+6\) and \(g(x)=2x\), what is \((f+g)(3)\)?

A) 9
B) 12
C) 15
D) 18

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\((f+g)(3)=f(3)+g(3)=9+6=15\).

Answer: C

Question 8

If \(f(x)=2x-1\) and \(g(x)=x^2\), what is \(f(g(3))\)?

F) 17
G) 15
H) 18
J) 35

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\(g(3)=9\), then \(f(9)=2(9)-1=17\).

Answer: F

Question 9

The graph shown approaches the horizontal line \(y=1\) as \(x\) decreases and rises rapidly as \(x\) increases. Which function could produce this graph?

A) \(y=2^x+1\)
B) \(y=2^x-1\)
C) \(y=-2^x+1\)
D) \(y=x^2+1\)

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\(y=2^x+1\) is increasing and has horizontal asymptote \(y=1\).

Answer: A

Question 10

If \(f(x)=5x+2\), which equation defines \(f^{-1}(x)\)?

F) \(f^{-1}(x)=5x-2\)
G) \(f^{-1}(x)=\frac{x+2}{5}\)
H) \(f^{-1}(x)=\frac{x-2}{5}\)
J) \(f^{-1}(x)=\frac{5}{x-2}\)

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Let \(y=5x+2\): swap \(x\) and \(y\).

\(x=5y+2\).

Solving gives \(y=\frac{x-2}{5}\).

Answer: H

Question 11

A function satisfies \(f(-x)=f(x)\) for every \(x\) in its domain. Which statement must be true?

A) \(f\) is odd.
B) \(f\) is even.
C) \(f\) is one-to-one.
D) \(f\) has no zeros.

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The condition \(f(-x)=f(x)\) is the definition of an even function.

Answer: B

Question 12

The graph of \(g\) is obtained from the graph of \(f\) by shifting it 2 units to the right and 1 unit down, as shown. Which equation relates the functions?

F) \(g(x)=f(x+2)+1\)
G) \(g(x)=f(x-2)+1\)
H) \(g(x)=f(x+2)-1\)
J) \(g(x)=f(x-2)-1\)

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A shift 2 units right replaces \(x\) with \(x-2\), and a shift 1 unit down subtracts 1: \(g(x)=f(x-2)-1\).

Answer: J

Question 13

The population of a culture is modeled by \(P(t)=800(1.12)^t\), where \(t\) is in hours. What does the number 1.12 represent?

A) An initial population of 112
B) A 12% decrease per hour
C) A 12% increase per hour
D) An increase of 112 organisms per hour

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A growth factor of \(1.12=1+0.12\) represents a 12% increase each hour.

Answer: C

Question 14

The piecewise graph of \(f\) is shown. At \(x=0\), which value is \(f(0)\)?

F) -1
G) 0
H) 1
J) 2

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At \(x=0\), the filled point is at \(y=-1\).

The open point belongs to the other piece and is not included.

Answer: F

Question 15

If \(f(x)=\frac{x-4}{x+1}\), which value is excluded from the domain of \(f\)?

A) -4
B) -1
C) 1
D) 4

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The denominator cannot equal 0.

Since \(x+1=0\) at \(x=-1\), that value is excluded.

Answer: B

Question 16

For \(f(x)=3^{x+1}\), which expression is equal to \(f(x+2)\)?

F) \(3^{x+2}\)
G) \(3^{x+3}\)
H) \(9^{x+1}\)
J) \(27\cdot 3^x\)

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\(f(x+2)=3^{(x+2)+1}=3^{x+3}=27\cdot 3^x\).

Answer: J

Question 17

Functions \(f\) and \(g\) are defined by \(f(x)=x^2-1\) and \(g(x)=\sqrt{x+1}\). For which \(x\) is \(g(f(x))\) defined?

A) All real \(x\)
B) \(x\ge 0\)
C) \(x\le 0\)
D) \(x\ne 0\)

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\(g(f(x))=\sqrt{(x^2-1)+1}=\sqrt{x^2}=|x|\), which is defined for every real \(x\).

Answer: A

Question 18

The functions \(f\) and \(g\) are defined by \(f(x)=\begin{cases}x+4,&x<0\\x^2,&x\ge 0\end{cases}\) and \(g(x)=\begin{cases}2x,&x\le 2\\x-1,&x>2\end{cases}\). What is \(f(g(-1))\)?

F) -4
G) 2
H) 4
J) 6

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\(g(-1)=2(-1)=-2\) because \(-1\le 2\).

Then \(f(-2)=-2+4=2\) because \(-2<0\).

Answer: G

Question 19

The graph has midline \(y=0\), amplitude 2, period 12, and crosses the midline upward at \(x=0\). Which equation could model the graph?

A) \(y=2\cos\left(\frac{\pi x}{6}\right)\)
B) \(y=\sin\left(\frac{\pi x}{6}\right)\)
C) \(y=2\sin\left(\frac{\pi x}{6}\right)\)
D) \(y=2\sin\left(\frac{\pi x}{12}\right)\)

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For \(y=A\sin(Bx)\), the amplitude is \(|A|\) and the period is \(\frac{2\pi}{B}\).

A period of 12 gives \(B=\frac{\pi}{6}\).

The upward midline crossing at \(x=0\) matches sine: \(y=2\sin\left(\frac{\pi x}{6}\right)\).

Answer: C

Question 20

If \(\log_2(x-1)=4\), what is the value of \(x\)?

F) 8
G) 15
H) 16
J) 17

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\(\log_2(x-1)=4\) means \(x-1=2^4=16\), so \(x=17\).

Answer: J