ACCUPLACER Advanced Algebra & Functions Quiz: Piecewise Functions

20 questions · exam conditions

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Piecewise FunctionsQuestion 1 of 20

For what value of the constant $c$ is the function $f(x) = \begin{cases} cx + 1 & \text{if } x \le 3 \\ x^2 - c & \text{if } x > 3 \end{cases}$ continuous at $x = 3$ ?

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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Piecewise Functions

Practice Piecewise Functions in ACCUPLACER Advanced Algebra & Functions 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 Piecewise Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

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

For what value of the constant $c$ is the function $f(x) = \begin{cases} cx + 1 & \text{if } x \le 3 \\ x^2 - c & \text{if } x > 3 \end{cases}$ continuous at $x = 3$ ?

0
2 (correct answer)
4
5

Explanation: For the function to be continuous at

x = 3

, the values of the two pieces must be equal at

x = 3

. We set the expression for

x \le 3

equal to the expression for

x > 3

and substitute

x = 3

. This gives the equation

c(3) + 1 = (3)^2 - c

. Simplifying, we get

3c + 1 = 9 - c

. Adding

c

to both sides gives

4c + 1 = 9

. Subtracting 1 gives

4c = 8

. Dividing by 4 yields

c = 2

Question 2

What is the range of the function $f(x) = \begin{cases} x & \text{if } x < 1 \\ 3 & \text{if } x = 1 \\ (x-1)^2 + 1 & \text{if } x > 1 \end{cases}$ ?

$(-\infty, \infty)$
$(-\infty, 1) \cup (1, \infty)$
$(-\infty, 1) \cup (1, \infty) \cup \{3\}$ (correct answer)
$(1, \infty)$

Explanation: We analyze the range of each piece. For

x < 1

f(x) = x

covers all values in the interval

(-\infty, 1)

. For

x = 1

f(x) = 3

, so the single value

3

is in the range. For

x > 1

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

is a parabola with its vertex at

(1, 1)

. Since

x > 1

, this piece covers all values in the interval

(1, \infty)

. Combining these parts, the total range is

(-\infty, 1) \cup (1, \infty) \cup \{3\}

Question 3

Consider the function $g(x) = \begin{cases} 2x + 7 & \text{if } x \le -1 \\ x^2 + 4 & \text{if } x > -1 \end{cases}$ . What is the average rate of change of $g(x)$ over the interval $[-3, 2]$ ?

-1
$-\frac{2}{5}$
$\frac{7}{5}$ (correct answer)
2

Explanation: The average rate of change is given by the formula

\frac{g(b) - g(a)}{b - a}

. Here,

a = -3

and

b = 2

. First, find

g(-3)

. Since

-3 \le -1

, use the first piece:

g(-3) = 2(-3) + 7 = 1

. Next, find

g(2)

. Since

2 > -1

, use the second piece:

g(2) = 2^2 + 4 = 8

. Now, calculate the average rate of change:

\frac{g(2) - g(-3)}{2 - (-3)} = \frac{8 - 1}{2 + 3} = \frac{7}{5}

Question 4

Which of the following piecewise functions is equivalent to $f(x) = |x - 3| + 2x$ ?

$f(x) = \begin{cases} x - 3 & \text{if } x < 3 \\ 3x - 3 & \text{if } x \ge 3 \end{cases}$
$f(x) = \begin{cases} x + 3 & \text{if } x < 3 \\ x - 3 & \text{if } x \ge 3 \end{cases}$
$f(x) = \begin{cases} 3x - 3 & \text{if } x < 3 \\ x + 3 & \text{if } x \ge 3 \end{cases}$
$f(x) = \begin{cases} x + 3 & \text{if } x < 3 \\ 3x - 3 & \text{if } x \ge 3 \end{cases}$ (correct answer)

Explanation: The absolute value function

|x - 3|

can be written as a piecewise function: it is

-(x - 3)

3 - x

when

x < 3

, and it is

x - 3

when

x \ge 3

. We must add

2x

to each piece. For

x < 3

f(x) = (3 - x) + 2x = x + 3

. For

x \ge 3

f(x) = (x - 3) + 2x = 3x - 3

. Combining these gives the correct piecewise definition.

Question 5

Let $f(x) = \begin{cases} x + 3 & \text{if } x < 1 \\ 5 & \text{if } x \ge 1 \end{cases}$ and $g(x) = x^2+1$ . At which positive $x$ -value do the graphs of $f(x)$ and $g(x)$ intersect?

-2
-1
1
2 (correct answer)

Explanation: We must check for intersections on both intervals defined by

f(x)

. Case 1:

x < 1

. Set

x + 3 = x^2 + 1

. This gives

x^2 - x - 2 = 0

, which factors to

(x-2)(x+1) = 0

. The solutions are

x = 2

and

x = -1

. Only

x=-1

satisfies the condition

x < 1

. Case 2:

x \ge 1

. Set

5 = x^2 + 1

. This gives

x^2 = 4

, so

x = 2

x = -2

. Only

x = 2

satisfies the condition

x \ge 1

. The two intersection points are at

x=-1

and

x=2

. The question asks for the positive

x

-value, which is 2.

Question 6

Let $f(x) = \begin{cases} -x-1 & \text{if } x < -1 \\ x+1 & \text{if } x \ge -1 \end{cases}$ and $g(x) = 2$ . What is the value of $f(f(-2)) + g(3)$ ?

1
2
3
4 (correct answer)

Explanation: This is a multi-step evaluation. First, find the value of the inner function

f(-2)

. Since

-2 < -1

, we use the first piece:

f(-2) = -(-2) - 1 = 2 - 1 = 1

. Now we must find

f(1)

. Since

1 \ge -1

, we use the second piece:

f(1) = 1 + 1 = 2

. So,

f(f(-2)) = 2

. Next, evaluate

g(3)

. The function

g(x)

is a constant function, so

g(3) = 2

. Finally, add the results:

f(f(-2)) + g(3) = 2 + 2 = 4

Question 7

A contractor's fee $F(h)$ in dollars for a job that takes $h$ hours is given by the function: $F(h) = \begin{cases} 50h & \text{if } 0 < h \le 10 \\ 500 + 75(h - 10) & \text{if } h > 10 \end{cases}$

A specific job had a total fee of $1100. How many hours did the job take?

8
16
18 (correct answer)
22

Explanation: First, check if the fee could be generated by the first rule. Set

50h = 1100

, which yields

h = 22

. This is not valid because the rule only applies for

h \le 10

. So, we must use the second rule. Set

500 + 75(h - 10) = 1100

. Subtract 500 from both sides:

75(h - 10) = 600

. Divide by 75:

h - 10 = 8

. Add 10 to both sides:

h = 18

. This is a valid solution because

18 > 10

Question 8

The value of a collectible is modeled by the function $V(t)$ , where $t$ is the number of years after 2010. $V(t) = \begin{cases} 500 + 100t & \text{if } 0 \le t \le 5 \\ 1000(1.04)^{t-5} & \text{if } t > 5 \end{cases}$ . What is the approximate value of the collectible in the year 2020?

$1000
$1217 (correct answer)
$1480
$1500

Explanation: The year 2020 corresponds to

t = 2020 - 2010 = 10

. Since

10 > 5

, we use the second piece:

V(t) = 1000(1.04)^{t-5}

. Substitute

t = 10

V(10) = 1000(1.04)^{10-5} = 1000(1.04)^5

. Calculating

(1.04)^5 \approx 1.217

, so

V(10) \approx 1000(1.217) = \

1217

. Choice C represents using

instead of

t-5

in the exponent, choice D uses the linear piece incorrectly, and choice A is the function value at

t=5$.

Question 9

The function $f(x)$ is defined as $f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & \text{if } x < 2 \\ ax - 1 & \text{if } x \ge 2 \end{cases}$ . For what value of $a$ is the function $f(x)$ continuous for all real numbers?

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

Explanation: For the function to be continuous at

x = 2

, the limit of the first piece as

x

approaches 2 must equal the value of the second piece at

x = 2

. First, find the limit:

\lim_{x\to 2^-} \frac{x^2 - 4}{x - 2} = \lim_{x\to 2^-} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x\to 2^-} (x + 2) = 4

. Now, set the value of the second piece at

x = 2

equal to this limit:

a(2) - 1 = 4

. This simplifies to

2a = 5

, so

a = \frac{5}{2}

Question 10

A company's shipping costs are as follows: a flat fee of $15 for any order weighing up to 5 pounds. For orders weighing more than 5 pounds, the cost is the $15 flat fee plus an additional $2 per pound for the weight exceeding 5 pounds. Which function $C(w)$ correctly models the shipping cost for an order weighing $w$ pounds, where $w > 0$ ?

$C(w) = \begin{cases} 15 & \text{if } 0 < w \le 5 \\ 2w & \text{if } w > 5 \end{cases}$
$C(w) = \begin{cases} 15 & \text{if } 0 < w \le 5 \\ 15+2w & \text{if } w > 5 \end{cases}$
$C(w) = \begin{cases} 15 & \text{if } 0 < w \le 5 \\ 2w+5 & \text{if } w > 5 \end{cases}$ (correct answer)
$C(w) = 15 + 2w$

Explanation: For weights

w

such that

0 < w \le 5

, the cost is a flat $15. For weights

w > 5

, the cost is the base $15 plus $2 for each pound over 5. The weight exceeding 5 pounds is

w - 5

. So the cost for

w > 5

15 + 2(w - 5)

. Distributing the 2 gives

15 + 2w - 10

, which simplifies to

2w + 5

. Therefore, the correct piecewise function is

C(w) = \begin{cases} 15 & \text{if } 0 < w \le 5 \\ 2w+5 & \text{if } w > 5 \end{cases}

Question 11

Let the function $f(x)$ be defined by $f(x) = \begin{cases} 3x + 10 & \text{if } x < -2 \\ 4 & \text{if } -2 \le x < 3 \\ x^2 - 5 & \text{if } x \ge 3 \end{cases}$ . For which value of $x$ is $f(x) = 11$ ?

$\frac{1}{3}$
-4
4 (correct answer)
-4 and 4

Explanation: We must test each piece of the function. For the first piece, set

3x + 10 = 11

, which gives

3x = 1

x = \frac{1}{3}

. However, this piece is only defined for

x < -2

, and

\frac{1}{3}

is not less than -2, so this is not a solution. For the second piece,

4 = 11

has no solution. For the third piece, set

x^2 - 5 = 11

, which gives

x^2 = 16

, so

x = 4

x = -4

. This piece is defined for

x \ge 3

. The value

x = 4

satisfies this condition, but

x = -4

does not. Therefore, the only solution is

x = 4

Question 12

The function $h(t)$ models the height of a drone in meters at time $t$ in seconds: $h(t) = \begin{cases} 5t & \text{if } 0 \le t \le 10 \\ 50 & \text{if } 10 < t \le 20 \\ -2(t-20) + 50 & \text{if } 20 < t \le 45 \end{cases}$

Which statement accurately describes the drone's movement?

The drone ascends, hovers, and then descends at a slower rate than its ascent. (correct answer)
The drone ascends, hovers, and then descends at a faster rate than its ascent.
The drone ascends, descends, and then hovers.
The drone ascends, hovers, and then descends at the same rate as its ascent.

Explanation: We analyze the slope (rate of change) for each time interval. From

t=0

t=10

, the function is

h(t)=5t

, so the slope is 5. This represents an ascent at 5 m/s. From

t=10

t=20

, the function is

h(t)=50

, a constant, so the slope is 0. This represents hovering. From

t=20

t=45

, the function is

h(t)=-2(t-20)+50 = -2t+90

, so the slope is -2. This represents a descent at a rate of 2 m/s. Since

2 < 5

, the drone descends at a slower rate than it ascends.

Question 13

The function $f(x)$ is defined as $f(x) = \begin{cases} x+k & \text{if } x < 2 \\ 3 & \text{if } x=2 \\ kx^2-7 & \text{if } x > 2 \end{cases}$ . The $\lim_{x\to 2} f(x)$ exists. What is the value of $k$ ?

1
3 (correct answer)
4
5

Explanation: For the limit at

x=2

to exist, the left-hand limit must equal the right-hand limit. The value of the function at

x=2

, which is 3, does not affect the existence of the limit itself. Left-hand limit:

\lim_{x\to 2^-} (x+k) = 2+k

. Right-hand limit:

\lim_{x\to 2^+} (kx^2-7) = k(2^2)-7 = 4k-7

. Set the limits equal:

2+k = 4k-7

. Subtract

k

from both sides:

2 = 3k-7

. Add 7 to both sides:

9 = 3k

. Divide by 3:

k=3

Question 14

The function $f(x)$ is defined as $f(x) = \begin{cases} x^2+a & \text{if } x < 1 \\ bx+2 & \text{if } x \ge 1 \end{cases}$ . The function is continuous at $x=1$ and $f(3) = 23$ . What is the value of $a$ ?

2
3
5
8 (correct answer)

Explanation: First, use

f(3) = 23

. Since

3 \ge 1

, we use the second piece:

b(3) + 2 = 23

, giving

3b = 21

, so

b = 7

. For continuity at

x=1

, the pieces must be equal:

x^2+a = bx+2

x=1

. Substituting

x=1

and

b=7

1^2+a = 7(1)+2

, which gives

1+a = 9

, so

a = 8

Question 15

A water tank's volume changes according to: $V(h) = \begin{cases} 50h & \text{if } 0 \leq h \leq 8 \\ 400 + 75(h-8) & \text{if } 8 < h \leq 12 \\ 700 & \text{if } h > 12 \end{cases}$ where $h$ is height in feet. At what height does the tank's volume first reach 550 cubic feet?

10 feet (correct answer)
10.5 feet
11 feet
Cannot be determined from given information

Explanation: Check each piece to find where V(h) = 550. First piece (0 ≤ h ≤ 8): 50h = 550 gives h = 11, but 11 > 8, so this piece doesn't reach 550. Second piece (8 < h ≤ 12): 400 + 75(h-8) = 550, so 75(h-8) = 150, thus h-8 = 2, so h = 10. Since 8 < 10 ≤ 12, this is valid. Third piece gives V(h) = 700 > 550. Therefore, the tank first reaches 550 cubic feet at h = 10 feet. Choice B incorrectly solves 400 + 75(h-8) = 550 as h = 10.5. Choice C uses wrong arithmetic. Choice D incorrectly assumes the volume cannot be determined.

Question 16

A delivery service charges based on package weight using this piecewise function: $f(w) = \begin{cases} 8.50 & \text{if } 0 < w \leq 2 \\ 8.50 + 3.25(w-2) & \text{if } 2 < w \leq 10 \\ 34.50 + 4.75(w-10) & \text{if } w > 10 \end{cases}$ where $w$ is weight in pounds. If a customer ships three packages weighing 1.5 lbs, 6 lbs, and 12 lbs respectively, what is the total shipping cost?

$46.00
$56.00
$52.00 (correct answer)
$48.25

Explanation: For the 1.5 lb package: Since 0 < 1.5 ≤ 2, use f(w) = 8.50, so cost = $8.50. For the 6 lb package: Since 2 < 6 ≤ 10, use f(w) = 8.50 + 3.25(6-2) = 8.50 + 3.25(4) = 8.50 + 13.00 = $21.50. For the 12 lb package: Since 12 > 10, use f(w) = 34.50 + 4.75(12-10) = 34.50 + 4.75(2) = 34.50 + 9.50 = $22.00. Total = 8.50 + 21.50 + 22.00 = $52.00. Choice A incorrectly uses the first piece for all packages. Choice B incorrectly calculates the third piece as 34.50 + 4.75(12) instead of 4.75(12-10). Choice D makes an arithmetic error in the second piece calculation.

Question 17

A parking garage uses this rate structure: $R(t) = \begin{cases} 5 & \text{if } 0 < t \leq 2 \\ 5 + 2\lceil t-2 \rceil & \text{if } t > 2 \end{cases}$ where $t$ is hours parked and $\lceil \cdot \rceil$ is the ceiling function. If someone parks for exactly 4.3 hours, how much more do they pay compared to parking for exactly 2.8 hours?

$2.00 (correct answer)
$4.00
$6.00
$3.00

Explanation: For 4.3 hours: Since t > 2, use R(t) = 5 + 2⌈4.3-2⌉ = 5 + 2⌈2.3⌉ = 5 + 2(3) = 5 + 6 = $11. For 2.8 hours: Since t > 2, use R(t) = 5 + 2⌈2.8-2⌉ = 5 + 2⌈0.8⌉ = 5 + 2(1) = 5 + 2 = $7. The difference is $11 - $7 = $2. Choice B incorrectly calculates ceiling functions as floor functions. Choice C uses the raw differences without the ceiling function. Choice D makes an error in applying the piecewise conditions.

Question 18

A cell phone plan charges according to: $C(m) = \begin{cases} 25 & \text{if } 0 \leq m \leq 500 \\ 25 + 0.10(m-500) & \text{if } 500 < m \leq 2000 \\ 175 + 0.05(m-2000) & \text{if } m > 2000 \end{cases}$ where $m$ is minutes used. What is the average cost per minute for someone who uses exactly 1800 minutes?

$0.092 per minute
$0.089 per minute (correct answer)
$0.096 per minute
$0.100 per minute

Explanation: For m = 1800 minutes, since 500 < 1800 ≤ 2000, use the second piece: C(1800) = 25 + 0.10(1800-500) = 25 + 0.10(1300) = 25 + 130 = $155. Average cost per minute = $155/1800 = $0.08611... ≈ $0.089 per minute. Choice A incorrectly calculates using the third piece formula. Choice C uses an error in the arithmetic (155/1800). Choice D incorrectly uses just the overage rate of $0.10 without accounting for the base fee.

Question 19

A manufacturing company's daily production cost follows: $P(u) = \begin{cases} 1000 + 50u & \text{if } 0 \leq u \leq 40 \\ 3000 + 25u & \text{if } 40 < u \leq 80 \\ 5000 + 10u & \text{if } u > 80 \end{cases}$ where $u$ is units produced. The company wants to minimize cost per unit. Which production level achieves this?

40 units
80 units
60 units
100 units (correct answer)

Explanation: When you encounter a piecewise cost function and need to minimize cost per unit, you must calculate the cost per unit (total cost divided by units) at critical points where the function changes behavior. Let's evaluate cost per unit at each boundary and test point. For any production level

u

, cost per unit equals

\frac{P(u)}{u}

. At 40 units:

\frac{P(40)}{40} = \frac{1000 + 50(40)}{40} = \frac{3000}{40} = 75

per unit At 80 units:

\frac{P(80)}{80} = \frac{3000 + 25(80)}{80} = \frac{5000}{80} = 62.5

per unit At 100 units:

\frac{P(100)}{100} = \frac{5000 + 10(100)}{100} = \frac{6000}{100} = 60

per unit For choice C (60 units), we use the second piece:

\frac{P(60)}{60} = \frac{3000 + 25(60)}{60} = \frac{4500}{60} = 75

per unit Choice A (40 units) gives a cost per unit of $75, which is higher than other options. Choice B (80 units) yields $62.50 per unit, better than A but not optimal. Choice C (60 units) also gives $75 per unit, matching A's inefficiency. Choice D (100 units) achieves the lowest cost per unit at $60. The pattern here shows that as production increases beyond 80 units, the fixed cost component becomes spread over more units while the variable cost per unit drops to just $10, creating the most efficient production level. Study tip: With piecewise cost functions, always check the cost per unit at boundary points and test higher production levels—economies of scale often make higher production more efficient per unit.

Question 20

A function $g(x)$ is defined as: $g(x) = \begin{cases} x^2 - 4 & \text{if } x < 0 \\ 2x + 1 & \text{if } 0 \leq x < 3 \\ -x + 10 & \text{if } x \geq 3 \end{cases}$ What is the value of $g(3) - g(-2) + g(1.5)$ ?

7
4
8
11 (correct answer)

Explanation: When you encounter a piecewise function, you need to determine which piece (or rule) applies to each input value based on the given conditions. Let's evaluate each term in

g(3) - g(-2) + g(1.5)

: For

g(3)

: Since

x = 3

and

3 \geq 3

, we use the third piece:

g(3) = -3 + 10 = 7

For

g(-2)

: Since

x = -2

and

-2 < 0

, we use the first piece:

g(-2) = (-2)^2 - 4 = 4 - 4 = 0

For

g(1.5)

: Since

x = 1.5

and

0 \leq 1.5 < 3

, we use the second piece:

g(1.5) = 2(1.5) + 1 = 3 + 1 = 4

Therefore:

g(3) - g(-2) + g(1.5) = 7 - 0 + 4 = 11

Choice A (7) represents just the value of

g(3)

alone, missing the other terms. Choice B (4) could result from calculating only

g(1.5)

or from arithmetic errors in combining the terms. Choice C (8) might occur if you miscalculated one of the function values or made an error in the final arithmetic. Choice D (11) correctly accounts for all three function evaluations and their proper combination. Study tip: Always pay careful attention to the boundary conditions in piecewise functions. Notice that

x = 3

uses the third piece because of the "greater than or equal to" condition (

x \geq 3

), while

x = 0

would use the second piece. Double-check which inequality symbols include the boundary point.