College Algebra Quiz: Piecewise Functions Evaluate And Graph

5 questions · exam conditions

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Piecewise Functions Evaluate And GraphQuestion 1 of 5

A piecewise function is defined as $g(x) = \begin{cases} 2x + 1 & \text{if } x \leq -1 \\ x^2 & \text{if } -1 < x < 3 \\ 5 - x & \text{if } x \geq 3 \end{cases}$ . For which value(s) of $x$ does $g(x) = 4$ ?

x = -2

only

x = 2

only

x = 1

and

x = -2

x = 2

and

x = 1

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College Algebra Quiz

College Algebra Quiz: Piecewise Functions Evaluate And Graph

Practice Piecewise Functions Evaluate And Graph in College Algebra 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 Evaluate And Graph, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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

$x = -2$ only
$x = 2$ only (correct answer)
$x = 1$ and $x = -2$
$x = 2$ and $x = 1$

Explanation: We need to solve

g(x) = 4

in each piece. For

x \leq -1

2x + 1 = 4

, so

x = 1.5

. But

1.5 > -1

, so this is not valid. For

-1 < x < 3

x^2 = 4

, so

x = \pm 2

. Since we need

-1 < x < 3

, only

x = 2

works (

x = -2

doesn't satisfy

-2 > -1

). For

x \geq 3

5 - x = 4

, so

x = 1

. But

1 < 3

, so this is not valid. Therefore, only

x = 2

satisfies

g(x) = 4

Question 2

The piecewise function $s(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x \neq 3 \\ k & \text{if } x = 3 \end{cases}$ can be made continuous by choosing the appropriate value of $k$ . However, if $k = 8$ , what type of discontinuity occurs at $x = 3$ ?

No discontinuity; the function is continuous at $x = 3$
A removable discontinuity because the limit exists but differs from the function value (correct answer)
An infinite discontinuity because the function approaches infinity
A jump discontinuity because the left and right limits are different

Explanation: First, we find the limit as

x \to 3

. For

x \neq 3

\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3

. Therefore,

\lim_{x \to 3} s(x) = \lim_{x \to 3} (x + 3) = 6

. Since

s(3) = k = 8

and

\lim_{x \to 3} s(x) = 6 \neq 8

, there is a discontinuity. It's removable because the limit exists (equals 6) but doesn't equal the function value (8). The discontinuity could be removed by setting

k = 6

Question 3

A parking garage charges according to the function $C(t) = \begin{cases} 3 & \text{if } 0 < t \leq 2 \\ 3 + 2(t - 2) & \text{if } 2 < t \leq 6 \\ 11 & \text{if } t > 6 \end{cases}$ , where $C(t)$ is the cost in dollars and $t$ is time in hours. If someone pays exactly $7, what is the maximum time they could have parked?

$5$ hours
$4.5$ hours
$3.5$ hours
$4$ hours (correct answer)

Explanation: When you encounter piecewise functions like this parking garage pricing model, you need to determine which piece of the function applies to your situation, then work within that constraint. Since someone paid exactly $7, you need to find where $

C(t) = 7

$ and determine the maximum possible time. Let's check each piece systematically. For

0 < t \leq 2

: The cost is flat at

3, so this piece can't give us

7. For

2 < t \leq 6

: Here

C(t) = 3 + 2(t - 2)

. Setting this equal to 7:

7 = 3 + 2(t - 2)

4 = 2(t - 2)

2 = t - 2

t = 4

Since

is in the valid range

2 < t \leq 6

, this works. For

t > 6

: The cost is flat at

11, which exceeds

7, so this piece is impossible. Therefore, the maximum time for $7 is exactly 4 hours. Looking at the wrong answers: Choice A (

5

hours) would actually cost

3 + 2(5-2) = 9

dollars, which is too expensive. Choice B (

4.5

hours) would cost

3 + 2(4.5-2) = 8

dollars, also too much. Choice C (

3.5

hours) would cost

3 + 2(3.5-2) = 6

dollars, which is less than $7 but not the maximum time possible. Study tip: With piecewise functions, always identify which piece applies to your constraint first, then solve within that domain. Don't forget to verify your answer falls within the piece's specified range.

Question 4

Consider the piecewise function $k(x) = \begin{cases} |x + 1| & \text{if } x \leq 2 \\ 3x - 5 & \text{if } x > 2 \end{cases}$ . What is the range of $k(x)$ for $x \in [-3, 4]$ ?

$[0, 7]$ (correct answer)
$[0, 3] \cup [1, 7]$
$[1, 7]$
$[0, 3] \cup (1, 7]$

Explanation: For

x \in [-3, 2]

, we use

k(x) = |x + 1|

. The vertex of

|x + 1|

is at

x = -1

, where

k(-1) = 0

. At the endpoints:

k(-3) = |-3 + 1| = 2

and

k(2) = |2 + 1| = 3

. Since the absolute value function decreases from

x = -3

x = -1

then increases from

x = -1

x = 2

, the range on this interval is

[0, 3]

. For

x \in (2, 4]

, we use

k(x) = 3x - 5

. This is increasing, so

k(x)

approaches

3(2) - 5 = 1

x \to 2^+

and reaches

k(4) = 3(4) - 5 = 7

. The range on this interval is

(1, 7]

. Combining: the overall range is

[0, 3] \cup (1, 7] = [0, 7]

since these intervals overlap.

Question 5

Consider $$p(x) = \begin{cases} x^2 - 1 & \text{if } x \neq 3 \ 5 & \text{if } x = 3 \end{cases}

The function has a jump discontinuity at $x = 3$ because the left and right limits are different
The function is continuous everywhere because $p(3)$ is defined
The function has a removable discontinuity at $x = 3$ since $\lim_{x \to 3} p(x) \neq p(3)$ (correct answer)
The function is continuous at $x = 3$ because both $p(3)$ and $\lim_{x \to 3} p(x)$ exist

Explanation: When you encounter a piecewise function like this, you need to analyze continuity by checking three conditions at any point of interest: the function value exists, the limit exists, and they're equal. Let's examine what happens at

x = 3

. First, find

\lim_{x \to 3} p(x)

. Since the limit only cares about values near 3 (not at 3), you use the first piece:

\lim_{x \to 3} (x^2 - 1) = 3^2 - 1 = 8

. Next, find

p(3)

. The function explicitly defines

p(3) = 5

. Since

\lim_{x \to 3} p(x) = 8

but

p(3) = 5

, the limit and function value don't match. This creates a removable discontinuity—you could "fix" it by redefining

p(3) = 8

. Choice A is wrong because both left and right limits equal 8 (they approach the same value from both sides). Choice B misses the point—having

p(3)

defined isn't enough for continuity; it must equal the limit. Choice D incorrectly assumes that having both the limit and function value exist guarantees continuity, but they must be equal, not just both exist. Choice C correctly identifies this as a removable discontinuity because the limit exists but doesn't equal the function value. Study tip: For piecewise functions, always check continuity at the boundary points by computing the limit using the appropriate piece, then comparing it to the actual function value at that point.