Functions Domain Test

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Question 1

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

A) $0$
B) $1$
C) $4$
D) $10$

Show solution

Evaluate from the inside out.

First, find $g(1) = 2(1) - 1 = 1$.

Then find $f(g(1)) = f(1) = (1)^2 + 3(1) = 1 + 3 = 4$.

1. $0$: computed $f(0) = 0 + 0 = 0$ (evaluated $f$ at $0$ instead of at $g(1)$)
1. $1$: stopped after computing $g(1) = 1$ without applying $f$
1. $10$: computed $f(2) = 4 + 6 = 10$ (evaluated $f$ at the input $x=2$ rather than at $g(1) = 1$)

Answer: C

Question 2

What is the domain of $f(x) = \sqrt{2x - 8}$?

F) $x \geq 0$
G) $x \geq 4$
H) $x \geq 8$
J) All real numbers

Show solution

The expression under the square root must be non-negative:

\[2x - 8 \geq 0 \implies 2x \geq 8 \implies x \geq 4\]

1. $x \geq 0$: set $x \geq 0$ directly without solving the inequality (treated the variable as the constraint)
1. $x \geq 8$: divided by $1$ instead of $2$: $2x \geq 8 \implies x \geq 8$ (forgot to divide both sides)
1. All real numbers: ignored the square root restriction (valid for $f(x) = 2x - 8$ but not $\sqrt{2x-8}$)

Answer: G

Question 3

The graph of $y = f(x)$ is shown below. What is the value of $f(3)$?

A) $0$
B) $1$
C) $3$
D) $4$

Show solution

To find $f(3)$, locate $x = 3$ on the horizontal axis and read the corresponding $y$-value from the graph.

At $x = 3$, the graph reaches its highest visible point at $y = 4$.

Therefore $f(3) = 4$.

1. $0$: read $f(1) = 0$ (the $x$-intercept of the graph) instead of $f(3)$
1. $1$: read $f(5) = 1$ (the rightmost plotted point) instead of $f(3)$
1. $3$: confused the input $x = 3$ with the output value, reporting the input as the answer

Answer: D

Question 4

Which of the following functions is even?

F) $f(x) = x^3 - x$
G) $f(x) = x^2 + x$
H) $f(x) = x^4 - 3x^2 + 1$
J) $f(x) = x^3 + 2x$

Show solution

A function is even if $f(-x) = f(x)$ for all $x$ (symmetric about the $y$-axis). This requires all exponents to be even.

For H: $f(-x) = (-x)^4 - 3(-x)^2 + 1 = x^4 - 3x^2 + 1 = f(x)$ ✓ — all terms have even exponents.

1. $f(x) = x^3 - x$: $f(-x) = -x^3 + x = -(x^3 - x) = -f(x)$ — this is an odd function
1. $f(x) = x^2 + x$: $f(-x) = x^2 - x \neq f(x)$ and $\neq -f(x)$ — neither even nor odd
1. $f(x) = x^3 + 2x$: $f(-x) = -x^3 - 2x = -f(x)$ — this is an odd function

Answer: H

Question 5

The table below defines a function $f$.

$x$	$-2$	$0$	$1$	$3$	$5$
$f(x)$	$7$	$3$	$1$	$-1$	$-5$

What is the value of $f(f(0))$?

A) $3$
B) $1$
C) $-1$
D) $9$

Show solution

Step 1: From the table, $f(0) = 3$.

Step 2: Find $f(f(0)) = f(3)$. From the table, $f(3) = -1$.

Therefore $f(f(0)) = -1$.

1. $3$: reported $f(0) = 3$ (stopped after the first step without applying $f$ a second time)
1. $1$: read the value $f(1) = 1$ from the table instead of $f(3)$
1. $9$: computed $f(0)^2 = 3^2 = 9$ (squared the first result instead of composing)

Answer: C

Question 6

A taxi charges a flat fee of $\$3.00$ plus $\$1.50$ per mile. Which function models the total cost $C$, in dollars, for a trip of $m$ miles?

F) $C(m) = 1.50m$
G) $C(m) = 3.00m + 1.50$
H) $C(m) = 1.50m + 3.00$
J) $C(m) = 4.50m$

Show solution

Total cost = (per-mile rate × miles) + flat fee:

\[C(m) = 1.50m + 3.00\]

1. $1.50m$: included only the per-mile charge, omitting the flat fee
1. $3.00m + 1.50$: swapped the rate and the flat fee — charged $\$3.00$ per mile instead of $\$1.50$
1. $4.50m$: added the rate and flat fee together and applied the sum as a single per-mile rate

Answer: H

Question 7

The function $f(x) = -(x - 2)(x + 4)$ is a downward-opening parabola. What are the $x$-intercepts of its graph?

A) $x = -2$ and $x = 4$
B) $x = 2$ and $x = -4$
C) $x = 2$ and $x = 4$
D) $x = -2$ and $x = -4$

Show solution

Set $f(x) = 0$:

\[-(x - 2)(x + 4) = 0\]

The negative sign does not affect where the product equals zero. Set each factor to zero:

\[(x - 2) = 0 \implies x = 2 \qquad (x + 4) = 0 \implies x = -4\]

1. $x = -2$ and $x = 4$: flipped the signs of both roots (common sign error reading from factored form)
1. $x = 2$ and $x = 4$: correct sign on $x = 2$ but wrong sign on the other: read $x + 4 = 0$ as $x = 4$
1. $x = -2$ and $x = -4$: negated both roots, perhaps from the leading negative sign

Answer: B

Question 8

The graph of $y = f(x)$ is reflected across the $x$-axis and then shifted up 5 units. Which of the following represents the resulting function?

F) $y = f(x) + 5$
G) $y = f(x + 5)$
H) $y = -f(x) + 5$
J) $y = -f(x + 5)$

Show solution

Reflect across the $x$-axis: replace $f(x)$ with $-f(x)$.
Shift up 5 units: add 5 outside the function.

Applying both in order: $y = -f(x) + 5$.

1. $y = f(x) + 5$: applied the vertical shift but forgot the reflection
1. $y = f(x + 5)$: shifted horizontally (left 5) instead of reflecting and shifting vertically
1. $y = -f(x + 5)$: reflected correctly but shifted the input horizontally rather than shifting the output vertically

Answer: H

Question 9

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

A) $1$
B) $4$
C) $9$
D) $25$

Show solution

Evaluate inside out:

\[f(2) = 2(2) + 1 = 5\] \[g(f(2)) = g(5) = 5 - 3 = 2\] \[h(g(f(2))) = h(2) = 2^2 = 4\]

1. $1$: made an error computing $g(5) = 5 - 3 = 2$; perhaps computed $g(f(2)) = g(5) - 4 = 1$… or $h(1) = 1$ from $g(f(0))$
1. $9$: computed $h(3)$ — perhaps $g(5) = 5 - 2 = 3$ (subtracted $2$ instead of $3$): $h(3) = 9$
1. $25$: stopped after $f(2) = 5$ and applied $h$ directly: $h(5) = 25$ (skipped the $g$ step)

Answer: B

Question 10

What is the domain of $f(x) = \dfrac{x + 1}{x^2 - 5x + 6}$?

F) All real numbers except $x = -1$
G) All real numbers except $x = 2$
H) All real numbers except $x = 2$ and $x = 3$
J) All real numbers except $x = 5$ and $x = 6$

Show solution

The function is undefined when the denominator equals zero. Factor:

\[x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3\]

The domain is all real numbers except $x = 2$ and $x = 3$.

1. All real numbers except $x = -1$: set the numerator equal to zero instead of the denominator
1. All real numbers except $x = 2$: found one root of the denominator but missed $x = 3$
1. All real numbers except $x = 5$ and $x = 6$: read the coefficients $-5$ and $+6$ directly from the denominator without factoring

Answer: H

Question 11

What is the range of $f(x) = -2(x - 3)^2 + 7$?

A) $y \leq 7$
B) $y \geq 7$
C) $y \leq 3$
D) All real numbers

Show solution

The function is in vertex form $f(x) = a(x - h)^2 + k$ with $a = -2$, $h = 3$, $k = 7$.

Since $a = -2 < 0$, the parabola opens downward, so the vertex $(3, 7)$ is the maximum.

The output can equal $7$ (at the vertex) and decreases without bound as $x$ moves away from $3$.

Range: $y \leq 7$.

1. $y \geq 7$: concluded the vertex is a minimum (upward-opening), ignoring $a < 0$
1. $y \leq 3$: used the $x$-coordinate of the vertex ($h = 3$) as the range bound instead of $k = 7$
1. All real numbers: correct for a linear function but not for a downward-opening parabola with a maximum

Answer: A

Question 12

Which of the following functions is odd?

F) $f(x) = x^4 + 2$
G) $f(x) = x^3 - 2x$
H) $f(x) = x^2 + x$
J) $f(x) = x^2 - 4$

Show solution

A function is odd if $f(-x) = -f(x)$ for all $x$ (symmetric about the origin). This requires all exponents to be odd.

For G: $f(-x) = (-x)^3 - 2(-x) = -x^3 + 2x = -(x^3 - 2x) = -f(x)$ ✓

1. $f(x) = x^4 + 2$: $f(-x) = x^4 + 2 = f(x)$ — this is even (but the constant $+2$ would make it neither if the definition were applied strictly; actually $f(-x) = f(x)$ ✓ since $(-x)^4 = x^4$, so this IS even)
1. $f(x) = x^2 + x$: $f(-x) = x^2 - x$, which equals neither $f(x)$ nor $-f(x)$ — neither even nor odd
1. $f(x) = x^2 - 4$: $f(-x) = x^2 - 4 = f(x)$ — this is even

Answer: G

Question 13

The piecewise function $f$ is defined as:

\[f(x) = \begin{cases} x^2 - 1 & \text{if } x < 0 \\ 2x + 3 & \text{if } x \geq 0 \end{cases}\]

Which of the following correctly gives $f(-3)$ and $f(0)$?

A) $f(-3) = -10$ and $f(0) = -1$
B) $f(-3) = 8$ and $f(0) = 3$
C) $f(-3) = -3$ and $f(0) = 3$
D) $f(-3) = 8$ and $f(0) = -1$

Show solution

Since $-3 < 0$, use the first rule: $f(-3) = (-3)^2 - 1 = 9 - 1 = 8$.

Since $0 \geq 0$, use the second rule: $f(0) = 2(0) + 3 = 3$.

1. $f(-3) = -10$ and $f(0) = -1$: applied the second rule to $-3$: $2(-3) + 3 = -3$… or $(-3)^2 - 11 = -2$; and applied the first rule to $0$: $0^2 - 1 = -1$ (used wrong pieces for both)
1. $f(-3) = -3$ and $f(0) = 3$: applied the second rule to $-3$: $2(-3) + 3 = -3$ (used wrong piece for $x = -3$), but correctly evaluated $f(0) = 3$
1. $f(-3) = 8$ and $f(0) = -1$: correct for $f(-3)$ but applied the first rule to $0$: $0^2 - 1 = -1$ (used wrong piece for $x = 0$)

Answer: B

Question 14

A radioactive substance has an initial mass of 200 grams and decays at a rate of $12\%$ per year. Which function gives the mass $M$, in grams, after $t$ years?

F) $M(t) = 200(1.12)^t$
G) $M(t) = 200(0.12)^t$
H) $M(t) = 200 - 0.12t$
J) $M(t) = 200(0.88)^t$

Show solution

A $12\%$ decay per year means the substance retains $100\% - 12\% = 88\% = 0.88$ of its mass each year:

\[M(t) = 200(0.88)^t\]

1. $200(1.12)^t$: used $1 + 0.12 = 1.12$, modeling $12\%$ growth instead of $12\%$ decay
1. $200(0.12)^t$: used the decay rate $0.12$ as the base instead of the retention factor $0.88$; this would reduce the substance to nearly nothing after just 1 year ($200 \times 0.12 = 24$ g)
1. $200 - 0.12t$: modeled linear decay (subtracting $0.12$ grams per year), not exponential decay

Answer: J

Question 15

The function $f(x) = x^2 + 4x$ is restricted to $x \geq -2$. Which of the following gives $f^{-1}(x)$?

A) $f^{-1}(x) = \sqrt{x + 4} - 2$
B) $f^{-1}(x) = \sqrt{x - 4} + 2$
C) $f^{-1}(x) = \sqrt{x + 4} + 2$
D) $f^{-1}(x) = -\sqrt{x + 4} - 2$

Show solution

Complete the square to find the inverse. Set $y = x^2 + 4x$:

\[y = (x + 2)^2 - 4\]

Swap $x$ and $y$, then solve for $y$:

\[x = (y + 2)^2 - 4 \implies (y + 2)^2 = x + 4 \implies y + 2 = \sqrt{x + 4}\]

(Taking the positive root since $x \geq -2$ means $y + 2 \geq 0$.)

\[y = \sqrt{x + 4} - 2\]

So $f^{-1}(x) = \sqrt{x + 4} - 2$.

Verify: $f(f^{-1}(x)) = (\sqrt{x+4} - 2)^2 + 4(\sqrt{x+4} - 2) = (x+4 - 4\sqrt{x+4} + 4) + (4\sqrt{x+4} - 8) = x + 4 + 4 - 8 = x$ ✓

1. $\sqrt{x - 4} + 2$: sign error — used $-4$ inside the radical and $+2$ outside (both signs flipped)
1. $\sqrt{x + 4} + 2$: sign error on the constant — should subtract $2$, not add it: $y + 2 = \sqrt{x+4}$ gives $y = \sqrt{x+4} - 2$, not $+2$
1. $-\sqrt{x + 4} - 2$: took the negative root (valid for the branch $x < -2$ but not for the restricted domain $x \geq -2$)

Answer: A

Question 16

What is the solution to $\log_3(x + 5) + \log_3(x - 1) = 3$?

F) $x = 4$
G) $x = 4$ and $x = -8$
H) $x = -8$
J) $x = 3$

Show solution

Use the product rule: $\log_3[(x+5)(x-1)] = 3$.

Convert to exponential form: $(x+5)(x-1) = 3^3 = 27$.

Expand: $x^2 + 4x - 5 = 27 \implies x^2 + 4x - 32 = 0$.

Factor: $(x + 8)(x - 4) = 0 \implies x = -8$ or $x = 4$.

Check for extraneous solutions: the arguments of both logarithms must be positive.

$x = 4$: $\log_3(9) + \log_3(3) = 2 + 1 = 3$ ✓
$x = -8$: $\log_3(-3)$ is undefined — extraneous, rejected.

Only $x = 4$ is valid.

1. $x = 4$ and $x = -8$: found both algebraic solutions but failed to check for extraneous roots
1. $x = -8$: kept only the extraneous solution and discarded the valid one
1. $x = 3$: confused the base ($3$) with the solution — a common log-equation mistake

Answer: F

Question 17

What is the period of $f(x) = \sin\!\left(\dfrac{\pi x}{4}\right) + 2$?

A) $2$
B) $4$
C) $8$
D) $\pi$

Show solution

For $f(x) = \sin(Bx) + k$, the period is $\dfrac{2\pi}{B}$.

Here $B = \dfrac{\pi}{4}$, so:

\[\text{Period} = \frac{2\pi}{\pi/4} = 2\pi \times \frac{4}{\pi} = 8\]

The $+2$ vertical shift does not affect the period.

1. $2$: used $B$ as the period directly: $B = \frac{\pi}{4} \approx 0.785$… or computed $\frac{\pi}{4} \times \frac{8}{\pi} = 2$ (inverted the formula)
1. $4$: computed $\frac{\pi}{2B} = \frac{\pi}{2 \cdot \pi/4} = \frac{\pi}{\pi/2} = 2$… or read $B$ directly as the period ignoring $\pi$
1. $\pi$: computed $\frac{2\pi}{2} = \pi$ (used $B = 2$ instead of $B = \pi/4$)

Answer: C

Question 18

The function $g$ is defined as:

\[g(x) = \begin{cases} -x + 6 & \text{if } x < 2 \\ x^2 - 1 & \text{if } 2 \leq x < 5 \\ 3x - 10 & \text{if } x \geq 5 \end{cases}\]

What is the value of $g(g(5))$?

F) $4$
G) $5$
H) $1$
J) $3$

Show solution

Step 1: Find $g(5)$. Since $5 \geq 5$, use the third rule:

\[g(5) = 3(5) - 10 = 15 - 10 = 5\]

Step 2: Find $g(g(5)) = g(5)$. Since $5 \geq 5$, use the third rule again:

\[g(5) = 3(5) - 10 = 5\]

Therefore $g(g(5)) = 5$.

Note: $x = 5$ is a fixed point of $g$ — applying $g$ to $5$ returns $5$.

1. $4$: applied the second piece ($2 \leq x < 5$) to $x = 5$: $5^2 - 1 = 24$… or the first piece: $-5 + 6 = 1$; or computed $g(4) = 4^2 - 1 = 15$, then subtracted 11
1. $1$: applied the first rule to $x = 5$: $-5 + 6 = 1$ (used the wrong piece for $x = 5$)
1. $3$: arithmetic error in the third rule: $3(5) - 10 - 2 = 3$, or used $3(5)/5 = 3$

Answer: G

Question 19

Two functions are defined as $f(x) = 3^x$ and $g(x) = 3^{x+2}$. Which of the following correctly compares their graphs?

A) The graph of $g$ is the graph of $f$ shifted right 2 units.
B) The graph of $g$ is the graph of $f$ shifted left 2 units.
C) The graph of $g$ is the graph of $f$ stretched vertically by a factor of 2.
D) The graph of $g$ is the graph of $f$ shifted up 2 units.

Show solution

$g(x) = 3^{x+2} = f(x + 2)$.

Replacing $x$ with $x + 2$ inside the function shifts the graph left by 2 units.

Equivalently: $g(x) = 3^{x+2} = 3^x \cdot 3^2 = 9 \cdot f(x)$. This is also a vertical stretch by a factor of 9 — but among the answer choices, the horizontal shift interpretation is the more direct reading of the transformation form $f(x + 2)$.

1. Shifted right 2 units: $f(x - 2) = 3^{x-2}$ would shift right; $f(x + 2)$ shifts left
1. Stretched vertically by a factor of 2: the factor is $3^2 = 9$, not 2; and this describes a vertical stretch, not a horizontal shift
1. Shifted up 2 units: $f(x) + 2 = 3^x + 2$ would shift up; adding to the exponent shifts horizontally

Answer: B

Question 20

Starting with $y = f(x)$, a student applies these transformations in order: (1) horizontally compress by a factor of $3$ (replace $x$ with $3x$); (2) reflect across the $y$-axis; (3) shift down 4 units. What is the resulting function?

F) $y = f(-3x) - 4$
G) $y = f(3x) - 4$
H) $y = -f(3x) - 4$
J) $y = f(-3x) + 4$

Show solution

Apply each transformation in order:

Compress horizontally by factor 3: replace $x$ with $3x$ → $y = f(3x)$
Reflect across $y$-axis: replace $x$ with $-x$ → $y = f(3(-x)) = f(-3x)$
Shift down 4: subtract 4 outside → $y = f(-3x) - 4$

1. $y = f(3x) - 4$: applied the compression and vertical shift correctly but omitted the $y$-axis reflection (step 2)
1. $y = -f(3x) - 4$: reflected across the $x$-axis (negated the output) instead of the $y$-axis (negated the input)
1. $y = f(-3x) + 4$: correct reflection and compression, but shifted up instead of down

Answer: F