Algebra 3 Quiz: Piecewise Functions

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Piecewise FunctionsQuestion 1 of 12

A gym charges a one-time $\$50$ enrollment fee, $\$20$ per month for the first 6 months, and $\$35$ per month after that. Which function gives the total cost $C(m)$ after $m$ months?

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Algebra 3 Quiz

Algebra 3 Quiz: Piecewise Functions

Practice Piecewise Functions in Algebra 3 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 Algebra 3.

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

A gym charges a one-time $\$50$ enrollment fee, $\$20$ per month for the first 6 months, and $\$35$ per month after that. Which function gives the total cost $C(m)$ after $m$ months?

(correct answer)

Explanation: When you see a cost scenario with different rates after a certain time, model it as a piecewise function and compute the cumulative cost at the switch point. For the first 6 months, the total cost is

50+20m

. At

m=6

, that is

50+120=170

. For

m>6

, you have already paid the enrollment fee and 6 months at $20, so the cost is

170+35(m-6)=170+35m-210=35m-40

. Thus the correct piecewise form is

50+20m

for

0\le m\le6

, and

35m-40

for

m>6

. The option using

50+35m

for

m>6

charges $35 for all months; you must subtract the $90 overcharge from the first 6 months, giving

50+35m-90=35m-40

. The option using

170+35m

for

m>6

correctly adds the first-six-month total but then also charges $35 for those same months, double-counting them. The option using

120+35m

omits the $50 enrollment fee and still charges $35 for every month; it also fails to include

m=6

. A reliable strategy: write the second piece as

C(6)+35(m-6)

, where

C(6)

is the total paid through the switch, rather than just

C(6)+35m

. This prevents both double-counting and missing the initial fee.

Question 2

Let $f(x)=\begin{cases}2x,&x<0\\x^2+1,&0\le x<2\\x+3,&x\ge2\end{cases}$ . At which $x$ -value(s), if any, is $f$ discontinuous?

$x=0$ and $x=2$
$x=2$ only
$x=0$ only (correct answer)
Neither; $f$ is continuous everywhere.

Explanation: When you see a piecewise function and are asked about continuity, your instinct should be to focus only on the breakpoints — the

x

-values where the formula changes. At each breakpoint, continuity requires that the left-hand limit, the right-hand limit, and the function value all agree. Here, the breakpoints are

x=0

and

x=2

. Start at

x=0

. For

x<0

, the function is

2x

, so the left-hand limit is

2(0)=0

. For

0\le x<2

, the function is

x^2+1

, so the right-hand limit is

0^2+1=1

. Since

0 \neq 1

, the limit does not exist, and

f

is discontinuous at

x=0

. Now check

x=2

. For

0\le x<2

, the left-hand limit is

2^2+1=5

. For

x\ge 2

, the function is

x+3

, so the right-hand limit is

2+3=5

. The function value is also

f(2)=5

. All three match, so

f

is continuous at

x=2

. Therefore, the only discontinuity is at

x=0

. The choice saying "

x=2

only" is wrong because

x=2

is continuous. The choice saying "

x=0

and

x=2

" is wrong for the same reason — it includes a point that is actually continuous. The choice "Neither;

f

is continuous everywhere" misses the clear jump at

x=0

, where the left and right limits differ. As a strategy, always test each breakpoint separately. Don't assume a piecewise function is discontinuous everywhere or nowhere — compute the one-sided limits and compare them to the function value at each boundary. That single check will save you from every trap on this type of question.

Question 3

Let $f(x)=\begin{cases}x+3,&x<-1\\\\x^2-1,&-1\le x<2\\\\5-x,&x\ge2\end{cases}$ . What is the value of $f(-2)+f(1)+f(3)$ ?

$7$
$5$
$3$ (correct answer)
$9$

Explanation: Whenever you see a piecewise function and are asked for a sum of values, your first job is to choose the correct branch for each input. Mistakes usually happen when an input is plugged into the wrong piece or at the interval boundaries. For

f(-2)

, since

-2<-1

, use

x+3

-2+3=1

. For

f(1)

, the middle condition

-1\le x<2

applies, so

1^2-1=0

. For

f(3)

, use the last piece

5-x

5-3=2

. Thus

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

. If you got

5

, you likely used

x^2-1

for

-2

, incorrectly treating

-2

as though it belonged in the middle piece and getting

3

. If you got

7

, you probably used

x+3

for

1

, forgetting that

1

is in the

-1\le x<2

interval, which gives

4

instead of

0

. If you got

9

, you used

x^2-1

for

3

, ignoring the

x\ge2

rule and computing

8

instead of

2

. Each wrong answer comes from applying a function rule outside its allowed interval. Study tip: for piecewise questions, check each

x

-value against all interval conditions before substituting. Notice whether the inequality is strict or inclusive at endpoints, since that determines which piece owns boundary values like

x=-1

and

x=2

Question 4

A store applies a 15% discount at checkout to any item whose original price is over $\$50$ . Items priced at $\$50$ or less are not discounted. If the sale price of an item was $\$60$ , what was its original price?

$\$69.00$
$\$70.59$ (correct answer)
$\$400.00$
$\$51.00$

Explanation: When you see a discount problem, remember that a 15% discount means the customer pays 85% of the original price — not that you add 15% to the sale price to get back to the original. Here, the sale price was $60, and since an item must be over $50 to receive the discount, the original price must be the unknown that was reduced by 15%. So set up: $

0.85 \times \text{original} = 60

. Divide both sides by 0.85:

\text{original} = 60 \div 0.85 = 70.588\ldots

$, which rounds to $70.59. This is over $50, so the discount correctly applies. The other choices each come from a common misunderstanding. $69.00 is found by computing $

60 \times 1.15$$, adding 15% of the sale price back to the sale price; but 15% off $69 would be $58.65, not $60. $400.00 treats the $60 as if it were the 15% discount amount itself, since $

60 \div 0.15 = 400$$; however, $60 is the price paid, not the amount saved. $51.00 comes from taking 15% off $60, i.e. $

60 \times 0.85 = 51

$, which mistakenly treats the sale price as the original price; if the original were $51, the discount would apply, but the sale price would be $43.35. Strategy: in any percent decrease problem, identify the “whole” (original price) and the “part” (final price). Use $

\text{final} = \text{original} \times (1 -

\text{percent as decimal}$)$$, and divide when you know the final amount.

Question 5

Let $f(x)=\begin{cases}x+2,&x<1\\3,&1\le x<3\\4-x,&x\ge3\end{cases}$ . Which statement correctly describes the graph of $f$ ?

At $x=3$ , the graph is continuous because every real $x$ appears in one of the pieces.
At $x=3$ , there is an open circle at $(3,3)$ and an open circle at $(3,1)$ .
At $x=1$ , there is an open circle at $(1,3)$ because the first piece does not include $x=1$ .
At $x=3$ , there is an open circle at $(3,3)$ and a closed circle at $(3,1)$ . (correct answer)

Explanation: Whenever you see a piecewise function, your job is to check each boundary point carefully: plug the boundary into the neighboring rules and see which rule actually includes that

x

-value. A closed circle means the point is part of the graph; an open circle means the rule approaches that value but does not include it. At

x=3

, look at the two pieces around it. The rule

3

applies for

1\le x<3

, so it gives the value

3

but does not include

x=3

. That means there is an open circle at

(3,3)

. The rule

4-x

applies for

x\ge3

, so it gives

4-3=1

and does include

x=3

. That means there is a closed circle at

(3,1)

. So the correct description is: open circle at

(3,3)

, closed circle at

(3,1)

. Now the wrong answers. Saying "every real

x

appears in one of the pieces" is not enough for continuity—continuity requires the two sides to meet at the same point, and here they jump. Saying there are open circles at both

(3,3)

and

(3,1)

is wrong because the third piece includes

x=3

, so its endpoint is closed. The statement about

x=1

—that there is an open circle at

(1,3)

because the first piece excludes

1

—misses that the second piece

3

applies for

1\le x<3

, so

x=1

is actually included, giving a closed circle at

(1,3)

. On this exam, always mark boundary points by checking each adjacent inequality:

\le

\ge

means closed,

<

>

means open. That simple check prevents most piecewise-function mistakes.

Question 6

Let $f(x)=\begin{cases}2x^2-1,&x<2\\\\kx+3,&x\ge2\end{cases}$ . For what value of $k$ is $f$ continuous at $x=2$ ?

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

Explanation: Whenever you see a piecewise function and a question about continuity at the break point, your job is to make the two pieces meet at that x-value. Here the switch is at

x=2

. From the left,

f(x)=2x^2-1

, so the left-hand limit is:

\lim_{x\to2^-}(2x^2-1)=2(2)^2-1=8-1=7.

For

x\ge2

, the function is

kx+3

, so at

x=2

we have:

f(2)=k(2)+3=2k+3.

Since the second piece is continuous for

x\ge2

, this is also the right-hand limit. Continuity requires the left-hand limit, right-hand limit, and actual value to all agree:

7=2k+3.

Subtract

3

from both sides:

4=2k

, so

k=2

. If

k=5

, then

f(2)=10+3=13

, leaving a huge jump from

7

. If

k=\frac72

, then

f(2)=7+3=10

, not

7

; this choice traps you if you forget the

+3

and just solve

2k=7

. If

k=3

, then

f(2)=6+3=9

, which also fails to match the left-hand limit of

7

. Only

k=2

makes both sides meet at the same value. adenAs your final check, plug

k=2

back in:

2(2)+3=7

, exactly matching the left-side limit. On continuity questions, always write down the one-sided limit at the break point, then set it equal to the function's value there — and don't drop any constant from the piecewise formula.

Question 7

Let $f(x)=\begin{cases}2x,&x<3\\\\x^2-4,&x\ge3\end{cases}$ and $g(x)=x-1$ . What is the value of $g(f(5))$ ?

$9$
$12$
$21$
$20$ (correct answer)

Explanation: When you see a composition like

g(f(5))

, always work from the inside out: first evaluate

f(5)

, then plug that result into

g

. Here, because

5 \ge 3

, use the second piece of

f

f(5)=5^2-4=25-4=21

. Then apply

g(x)=x-1

g(21)=21-1=20

. The choice 9 comes from incorrectly using the first piece of

f

2x

, for

x=5

even though that piece only applies when

x<3

. That gives

f(5)=10

, and then

g(10)=9

. The choice 21 is the value of

f(5)

itself — it shows the trap of stopping after evaluating the inner function and forgetting to apply

g

. The choice 12 results from reversing the order of composition: computing

g(5)=4

first, then

f(4)=4^2-4=12

. But the notation

g(f(5))

clearly tells you to do

f

first. On exam day, remember two things: check the condition of the piecewise function before picking its formula, and always finish every nested evaluation — don't stop at the inner value. Working inside-out step by step will keep each part of the composition clear.

Question 8

Let $f(x)=\begin{cases}x^2,&x<0\\\\2x+1,&0\le x<3\\\\6-x,&x\ge3\end{cases}$ . How many real numbers $x$ satisfy $f(x)=4$ ?

$1$
$2$ (correct answer)
$3$
$4$

Explanation: When you see a piecewise function and are asked how many inputs satisfy an equation, treat each piece as its own mini-problem—but check its domain. For

x<0

, set

x^2=4

. The solutions are

x=2

and

x=-2

, but only

x=-2

lies in the domain

x<0

. So this branch contributes 1. For

0\le x<3

, set

2x+1=4

. Then

2x=3

, so

x=1.5

, which is in the domain. This branch contributes 1. For

x\ge3

, set

6-x=4

. Then

x=2

, but

2

is not in

x\ge3

, so this branch contributes 0. Thus the real numbers satisfying

f(x)=4

are

x=-2

and

x=1.5

: exactly 2 solutions. Why the other choices are traps: "1" misses one valid branch, often by forgetting that the first branch can produce a negative solution. "3" often adds

x=2

from the third piece without checking the domain, or counts the excluded

x=2

from the first branch. "4" typically counts

x=2

twice—from both

x^2=4

and

6-x=4

—while also including the two valid solutions; but

x=2

is excluded by both domains. Study tip: after solving each piece, always verify the solution against that piece's domain restriction. The domain is the real trap in piecewise-function equation questions.

Question 9

Let $f(x)=\begin{cases}x+1,&x<-1\\\\x^2,&-1\le x<2\\\\4-x,&x\ge2\end{cases}$ . What is the range of $f$ ?

$(-\infty,4]$
$[0,4)$
$(-\infty,\infty)$
$(-\infty,4)$ (correct answer)

Explanation: When you see a piecewise function, the range is the union of the outputs each piece produces on its own domain. Think of each branch as a separate "machine" with its own input interval, then combine all possible

y

-values. For

x<-1

, the rule

x+1

produces every value below

0

, because as

x

approaches

-1

from the left,

x+1

approaches

0

but never reaches it, and as

x

goes to

-\infty

, so does

x+1

. This branch contributes

(-\infty,0)

. For

-1\le x<2

, the rule

x^2

gives values from

0

up to but not including

4

0

is reached at

x=0

, and

4

is approached as

x

gets close to

2

, but

2

is not included. This branch contributes

[0,4)

. For

x\ge2

, the rule

4-x

gives

2

x=2

, then decreases without bound, contributing

(-\infty,2]

. Unioning these gives

(-\infty,0)\cup[0,4)\cup(-\infty,2]=(-\infty,4)

. The choice

(-\infty,4]

is wrong because

4

is never actually produced. The choice

[0,4)

is wrong because it ignores all negative outputs from the first and third branches. The choice

(-\infty,\infty)

is wrong because the function never produces any value

4

or larger. Study tip: for piecewise range questions, test the endpoints and the behavior at the extremes of each interval, then union the results carefully. Watch especially for open versus closed endpoints — that's where traps live.

Question 10

Let $f(x)=\begin{cases}3x+2,&x<-1\\\\x^2-4,&-1\le x<2\\\\5-x,&x\ge2\end{cases}$ . What is the value of $f(f(-1))$ ?

$-7$ (correct answer)
$-3$
$7$
$11$

Explanation: Whenever you see a piecewise function inside another function, the key is to work step by step from the inside out. Composite notation

f(f(-1))

means plug

-1

into

f

first, then take that output and plug it into

f

again. . First, evaluate the inner value:

f(-1)

. Because

-1

satisfies

-1 \le x < 2

, use the middle rule:

f(-1)=(-1)^2-4=1-4=-3

. .Now take that result,

-3

, and evaluate

f(-3)

. Since

-3<-1

, use the first rule:

f(-3)=3(-3)+2=-9+2=-7

. Therefore,

f(f(-1))=-7

. . The choice

-3

is the result of stopping after the first evaluation, confusing the inner value with the final composite value. The choice

7

could come from flipping both signs when computing the first rule, writing

3(3)-2=7

, which treats the input as positive and subtracts instead of adding. Similarly,

11

likely comes from computing

3(3)+2=11

, using a positive input for a negative piece — a sign error on the domain condition. And if you mistakenly think

-3

falls in the middle rule, you'd get

(-3)^2-4=5

,which isn't listed; among the options,

11

may also arise from

9+2

without considering the input sign at all. . A good study habit for piecewise function questions: always write down which piece applies at each step, and check the input against the domain intervals before substituting. Evaluating inside-out prevents the most common trap of mixing up the inner value with the final answer.

Question 11

An electric company charges $\$0.12$ per kWh for the first 500 kWh, $\$0.15$ per kWh for the next 500 kWh, and $\$0.20$ per kWh for any usage above 1000 kWh. What is the total charge for using 1200 kWh?

$\$144$
$\$175$ (correct answer)
$\$190$
$\$240$

Explanation: Whenever you see a tiered pricing question, your first instinct should be to break the usage into blocks and apply each rate only to the kWh inside that block. Don't just multiply total usage by one number. For 1200 kWh, split it: the first 500 kWh cost

500 \times 0.12 = \$60

. The next 500 kWh cost

500 \times 0.15 = \$75

. The remaining 200 kWh are above 1000, so they cost

200 \times 0.20 = \$40

. Adding the three blocks gives

60 + 75 + 40 = \$175

. Each wrong answer comes from a flawed shortcut. $144 treats all 1200 kWh at the lowest rate of

0.12, ignoring the higher-tier charges. **

190** incorrectly charges the first 1000 kWh at $0.15, as if the

0.15 rate applied to the whole first block rather than only the second 500. **

240** charges all 1200 kWh at the highest rate of $0.20, ignoring the cheaper introductory tiers entirely. Your study tip: for any "piecewise" or tiered charge problem, draw mental boundaries at the thresholds, then calculate each slab separately before adding. Check that the middle rate only touches the middle block, never the first block. That block-by-block habit will save you on tax, utility, and discount questions too.

Question 12

Let $f(x)=\begin{cases}2x+1,&x<\frac12\\\\x^2,&x\ge\frac12\end{cases}$ . What is the value of $f\left(\frac12\right)-f\left(-\frac12\right)$ ?

$2$
$0$
$\frac14$ (correct answer)
$\frac74$

Explanation: Whenever you see a piecewise function, your first job is to locate each input on the number line and decide which branch applies. Here the function changes at

x=\frac12

. For

f\left(\frac12\right)), the condition is

x\ge\frac12

, so you must use the second branch:

f\left(

\frac

12\right)=\left(\frac12\right)^2=\frac14

. For

f\left(-

\frac

12\right)), since

-\frac12<\frac12

, you use the first branch:

f\left(-\frac12\right)=2\left(-\frac12\right)+1=0

. Thus the difference is

\frac14-0=\frac14

, so the correct choice is

\frac14

. The trap choices come from misassigning branches. The choice

2

results from incorrectly using

2x+1

x=\frac12

and forgetting that the "greater than or equal" sign sends that input to the quadratic piece. The choice

0

can happen if you evaluate both inputs with the quadratic branch:

f\left(-\frac12\right)=\frac14

and

f\left(\frac12\right)=\frac14

, then subtract to get

0

; but

-\frac12

is less than

\frac12

, so it belongs to the linear piece. The choice

\frac74

comes from mixing both mistakes: using the linear branch at

\frac12

to get

2

, using the quadratic branch at

-\frac12

to get

\frac14

, and computing

2-\frac14

. In every case, the error is ignoring the inequality signs. boundary condition. On future piecewise questions, circle the inequalities before plugging in. Ask: Which side is inclusive? The boundary

x=\frac12

belongs to the quadratic branch because of the "

x\ge\frac12

" sign, so evaluate each input on its correct side before subtracting. This habit will save you on similar exam questions.