Special Functions: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

Advanced. This question goes beyond core Algebra I. It is not required by the course.

What is the value of $\sqrt{49}$ ?

Answer choices for question 1 A $7$ B $\pm 7$ C $-7$ D $24.5$
2

For $h(x) = \begin{cases} 3x - 2 & \text{if } x < 4, \\ 10 - x & \text{if } x \ge 4, \end{cases}$ find $h(4)$ .

Answer choices for question 2 A $10$ B $-6$ C both $6$ and $10$ D $6$
3

Advanced. This question goes beyond core Algebra I. It is not required by the course.

What is the domain of $f(x) = \dfrac{x + 5}{x^2 - 3x - 10}$ ?

Answer choices for question 3 A all real numbers except $-5$ and $2$ B all real numbers except $5$ , $-2$ , and $-5$ C all real numbers except $5$ and $-2$ D all real numbers
4

Solve $\lvert 2x - 7 \rvert = 9$ .

Answer choices for question 4 A $x = 8$ only B $x = 8$ or $x = -1$ C $x = 1$ or $x = -8$ D $x = 8$ or $x = -8$
5

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Evaluate $\lfloor -4.6 \rfloor$ .

Answer choices for question 5 A $-5$ B $-4$ C $-6$ D $4$
6

Advanced. This question goes beyond core Algebra I. It is not required by the course.

What is the domain of $f(x) = \sqrt{5 - x}$ ?

Answer choices for question 6 A $x \ge 5$ B $x < 5$ C all real numbers D $x \le 5$
7

Advanced. This question goes beyond core Algebra I. It is not required by the course.

What is the horizontal asymptote of $f(x) = \dfrac{4x^2 + 3}{2x^2 + 6}$ ?

Answer choices for question 7 A $y = \tfrac{1}{2}$ B $y = 2$ C $y = 0$ D no horizontal asymptote
8

Advanced. This question goes beyond core Algebra I. It is not required by the course.

What is the fractional part $\{-2.6\}$ , where $\{x\} = x - \lfloor x \rfloor$ ?

Answer choices for question 8 A $0.6$ B $-0.6$ C $0.4$ D $2.6$
9

Solve $\lvert x + 2 \rvert \le 5$ .

Answer choices for question 9 A $-7 \le x \le 3$ B $x \le -7$ or $x \ge 3$ C $-3 \le x \le 7$ D $x \le 3$
10

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For $f(x) = \begin{cases} \sqrt{2x - 1} & \text{if } x \ge 1, \\ x^2 & \text{if } x < 1, \end{cases}$ find $f(5)$ .

Answer choices for question 10 A $25$ B $3$ C $9$ D $\pm 3$
11

Advanced. This question goes beyond core Algebra I. It is not required by the course.

How does the graph of $f(x) = \dfrac{x^2 - 4x - 21}{x^2 - 49}$ behave at $x = 7$ ?

Answer choices for question 11 A a vertical asymptote at $x = 7$ B a hole at $(7, 0)$ C a vertical asymptote at $x = -7$ D a hole at $\left(7, \tfrac{5}{7}\right)$
12

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For which values of $x$ is $\lceil x \rceil = -4$ ?

Answer choices for question 12 A $-5 < x \le -4$ B $-4 \le x < -3$ C $-4 < x \le -3$ D $-5 \le x < -4$
13

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Simplify $\sqrt{x^2}$ so that it is correct for every real number $x$ .

Answer choices for question 13 A $x$ B $\pm x$ C $\lvert x \rvert$ D $x^2$
14

At the boundary $x = 3$ , what does the graph of $m(x) = \begin{cases} x^2 & \text{if } x \le 3, \\ 11 & \text{if } x > 3 \end{cases}$ do?

Answer choices for question 14 A open dot at $(3, 9)$ , closed dot at $(3, 11)$ ; the graph jumps B closed dot at $(3, 9)$ , open dot at $(3, 11)$ ; the graph jumps C closed dot at $(3, 9)$ , open dot at $(3, 11)$ ; the graph connects D a single closed dot at $(3, 11)$ ; the graph connects
15

Which of these equations has exactly one solution?

Answer choices for question 15 A $\lvert x - 4 \rvert = 7$ B $\lvert x - 4 \rvert = -7$ C $\lvert x - 4 \rvert = 4$ D $\lvert x - 4 \rvert = 0$
16

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A ferry carries $40$ cars per crossing. If $218$ cars are waiting, how many crossings are needed to carry them all?

Answer choices for question 16 A $5$ B $5.45$ C $6$ D $54.5$
17

Which piecewise rule is equal to $\lvert x + 5 \rvert$ ?

Answer choices for question 17 A $\begin{cases} x + 5 & \text{if } x \ge -5 \\ -x - 5 & \text{if } x < -5 \end{cases}$ B $\begin{cases} x + 5 & \text{if } x \ge 0 \\ -x - 5 & \text{if } x < 0 \end{cases}$ C $\begin{cases} -x - 5 & \text{if } x \ge -5 \\ x + 5 & \text{if } x < -5 \end{cases}$ D $\begin{cases} x - 5 & \text{if } x \ge 5 \\ 5 - x & \text{if } x < 5 \end{cases}$
18

Advanced. This question goes beyond core Algebra I. It is not required by the course.

At which input, if any, is $f(x) = \begin{cases} \dfrac{6}{x - 1} & \text{if } x < 0, \\ x + 2 & \text{if } x \ge 0 \end{cases}$ undefined?

Answer choices for question 18 A undefined at $x = 1$ B undefined at $x = 0$ C undefined at $x = -1$ D defined for every real number
19

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Which statement about a rational function and its asymptotes is always correct?

Answer choices for question 19 A It can never cross either kind of asymptote. B It can cross a vertical asymptote but never a horizontal one. C It can cross a horizontal asymptote but never a vertical one. D It can cross both kinds of asymptote.
20

What is the range of $p(x) = \begin{cases} x - 3 & \text{if } x \le 2, \\ 7 & \text{if } x > 2 \end{cases}$ ?

Answer choices for question 20 A $y \le -1$ B $y \le -1$ or $y = 7$ C all real numbers D $-1 \le y \le 7$

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Where you stand

Evaluate a principal (nonnegative) square root, determine the domain of an even-index radical, and solve a radical equation, checking every candidate against the original to reject extraneous roots. Review: Radical Functions
Solve absolute-value equations and inequalities, decide how many solutions an equation has from the sign of the constant, express an absolute value through its two-piece rule, and recognize $\sqrt{x^2} = \lvert x \rvert$ . Review: Absolute Value Equations and Graphs
Evaluate floors, ceilings, and fractional parts correctly on negative inputs, solve a floor or ceiling equation for its interval of inputs, and use a ceiling or floor to count. Review: Floor and Ceiling
Find a rational function's domain, distinguish a vertical asymptote from a hole, determine a horizontal asymptote by comparing degrees, and reason about when a graph can meet an asymptote. Review: Rational Functions
Evaluate a piecewise function by choosing the governing piece, graph it with the correct open and closed dots and jump-or-connect verdict, read its domain and range, and recognize the absolute value and the floor as piecewise functions. Review: Piecewise Functions

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The layered equation

Solve $\dfrac{|6-2x|}{4}+5=8$ .
1. Hint 1
  The bars report a distance from zero, so they must stand alone before that distance can be read off.
2. Hint 2
  Once the bars are alone and the right side is a positive number, the inside equals that number or its opposite.
Answer

$x=-3$ or $x=9$ .

Full solution

Subtracting $5$ from both sides and then multiplying by $4$ leaves the bars alone, giving
$|6-2x|=12.$
The right side is positive, so the inside is $12$ or $-12$ .

One case is
$\begin{aligned}6-2x&=12\\-2x&=6\\x&=-3\end{aligned}$
and the other is
$\begin{aligned}6-2x&=-12\\-2x&=-18\\x&=9\end{aligned}$

Substituting $x=-3$ gives $|6+6|/4+5=3+5=8$ , and substituting $x=9$ gives $|6-18|/4+5=3+5=8$ , so both values satisfy the original equation.

Answer

$x=-3$ or $x=9$ .

Key idea

An absolute value must stand alone before a positive right side can be split into its two cases.
Problem 1: Mark complete
Problem 2 The repeating beep

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A timer beeps every $7$ seconds, with its first beep $7$ seconds after it starts. How many beeps occur during the first $100$ seconds?
1. Hint 1
  A beep arrives only once a whole $7$ -second stretch has gone by, so count the whole stretches.
2. Hint 2
  Divide the elapsed time by the interval and take the greatest integer at or below that quotient.
Answer

$14$ beeps.

Full solution

The beeps land at $7$ , $14$ , $21$ seconds and so on, so the count is the number of complete $7$ -second stretches inside $100$ seconds.
Since $100/7=14.28\ldots$ , that count is
$\left\lfloor\frac{100}{7}\right\rfloor=14.$

Checking the ends, the fourteenth beep is at $14(7)=98$ seconds, inside the first $100$ , while the next would be at $105$ seconds, outside them.

Answer

$14$ beeps.

Key idea

A floor counts only the complete intervals that have finished, leaving the part-finished one uncounted.
Problem 2: Mark complete
Problem 3 The excluded inputs

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Find the domain of $f(x)=\dfrac{x+6}{x^3-4x}$ .
1. Hint 1
  Only the denominator can force an input out of a rational function's domain.
2. Hint 2
  Take out the common factor of the denominator first, then set each remaining factor equal to zero.
Answer

All real numbers except $-2$ , $0$ and $2$ .

Full solution

Only the zeros of the denominator restrict the inputs, so factor it completely:
$x^3-4x=x(x-2)(x+2).$

A product is zero exactly when one of its factors is, so the denominator vanishes at $-2$ , $0$ and $2$ , and those three inputs are excluded.

Every other real number leaves the denominator nonzero and is allowed.
The numerator's zero at $x=-6$ restricts nothing, since only a denominator can.

Answer

All real numbers except $-2$ , $0$ and $2$ .

Key idea

The domain of a rational rule leaves out exactly the zeros of its denominator, which a complete factoring reveals.
Problem 3: Mark complete
Problem 4 The unfinished endpoints

The figure shows three pieces of one function, but the endpoint circles at inputs $-1$ and $2$ have not yet been assigned open or closed status. The function must be defined for every input from $-3$ through $4$ , with $f(-1)=4$ and $f(2)=-2$ . Complete those markings and state whether the graph jumps or connects at each of the two boundaries. Also state the range of the completed graph.

Three pieces of one function, with the four boundary circles at $x=-1$ and $x=2$ left unfinished.

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative four to five and the vertical y-axis from negative three to five, with gridlines, tick marks and number labels at every whole number, the origin labeled 0, and arrowheads at both ends of each axis. Three separate straight pieces of one function are drawn. One rises from the point (negative three, 0) to the point (negative one, 2). One is horizontal at height 4, running from the point (negative one, 4) to the point (2, 4). One rises from the point (2, negative two) to the point (4, 0). The outer ends at (negative three, 0) and (4, 0) carry filled dots. The four inner ends carry hollow circles and are labeled (negative one, 2), (negative one, 4), (2, 4) and (2, negative two). A line of text under the grid reads: Boundary circles are unfinished; fill the ones that belong to the function.
1. Hint 1
  The specified output at each boundary identifies exactly which endpoint belongs to the graph.
2. Hint 2
  Leave the other endpoint at that input open, then compare the two heights.
Answer

Close $(-1,4)$ and $(2,-2)$ ; leave $(-1,2)$ and $(2,4)$ open. The graph jumps at both boundaries. Range $-2\le y<2$ or $y=4$ (equivalently $[-2,2)\cup\{4\}$ ).

Full solution

At input $-1$ , the required output $4$ closes $(-1,4)$ and leaves $(-1,2)$ open.
At input $2$ , the required output $-2$ closes $(2,-2)$ and leaves $(2,4)$ open.

The two heights at each boundary differ.
Thus each boundary is a jump, with no vertical segment joining the two heights.
All inputs in the stated domain now have one output.

The left and right sloping pieces together attain $-2\le y<2$ .
The middle piece adds only height $4$ , so the range is $-2\le y<2$ or $y=4$ , that is
$[-2,2)\cup\{4\}.$
Height $2$ is excluded because $(-1,2)$ is open and no other piece reaches it.

Answer

Close $(-1,4)$ and $(2,-2)$ ; leave $(-1,2)$ and $(2,4)$ open. The graph jumps at both boundaries. Range $-2\le y<2$ or $y=4$ (equivalently $[-2,2)\cup\{4\}$ ).

Key idea

Boundary output records determine endpoint ownership, while the two neighboring heights determine whether the pieces join.
Problem 4: Mark complete
Problem 5 The matching output

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $f(x)=\sqrt{3x+1}$ for $0\le x\le5$ , and $f(x)=x-1$ for $x>5$ . Find every real input satisfying $f(x)=x-1$ .
1. Hint 1
  Solve the requested equation within each piece and retain its input condition.
2. Hint 2
  The root is nonnegative, and squaring its equation can add a candidate with the wrong sign.
Answer

Every $x\ge5$ .

Full solution

For $x>5$ , the second rule is already $x-1$ , so every such input works.
For $0\le x\le5$ , the root piece governs and equality requires
$\sqrt{3x+1}=x-1.$

Squaring gives $3x+1=x^2-2x+1$ , and collecting leaves
$x^2-5x=0.$
Its factors $x$ and $x-5$ give the candidates $0$ and $5$ , both inside this piece's interval.

At $0$ , the two sides are $1$ and $-1$ , so that candidate is extraneous.
At $5$ , both sides are $4$ , so it works.
Combining it with the second piece gives every $x\ge5$ .

Answer

Every $x\ge5$ .

Key idea

A piecewise equation combines solutions from each governing rule after checking both domain conditions and any irreversible algebra.
Problem 5: Mark complete
Problem 6 The shared target

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Define $F(x)=\lfloor x\rfloor$ for $x<0$ and $F(x)=\lceil x\rceil$ for $x\ge0$ . Find every input whose output is $0$ .
1. Hint 1
  Each piece may be tested on its own, but only on the inputs its own condition claims.
2. Hint 2
  Translate each bracket equality into its interval of inputs, then keep only the part of that interval the piece actually governs.
Answer

$x=0$ only.

Full solution

On the negative inputs the floor piece governs, and the floor equals $0$ exactly on
$0\le x<1.$
No input of that step is negative, so this piece produces nothing.

On the inputs $x\ge0$ the ceiling piece governs, and the ceiling equals $0$ exactly on
$-1<x\le0.$
The only input of that step which is also nonnegative is $x=0$ .

There $F(0)=\lceil0\rceil=0$ , so the single solution is $x=0$ .

Answer

$x=0$ only.

Key idea

A bracket equality answers with a whole step of inputs, and a piecewise rule keeps only the part of that step its own condition claims.
Problem 6: Mark complete
Problem 7 The two instructions

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $F(x)=(x^2+5x+6)/(x+2)$ for $x<1$ , and $F(x)=1/(x-4)$ for $x\ge1$ . Find its domain, every hole, and every vertical asymptote.
1. Hint 1
  Each denominator matters only on the inputs governed by its own piece.
2. Hint 2
  Cancel the first piece on its allowed nonzero denominator, then inspect the remaining forbidden inputs.
Answer

Domain: all real $x$ except $-2,4$ ; hole $(-2,1)$ ; vertical asymptote $x=4$ .

Full solution

In the first piece, $x=-2$ is excluded and lies inside its interval.
For $x\ne-2$ , factoring and canceling give
$\frac{x^2+5x+6}{x+2}=x+3.$
The reduced rule has value $1$ at $-2$ , so the missing point is $(-2,1)$ , a hole.

The second piece excludes $x=4$ , which belongs to its interval, and its uncanceled denominator produces the vertical asymptote $x=4$ .
The boundary $1$ is allowed by the second piece and returns $-1/3$ .

All other real inputs have one defined governing formula, so the domain excludes exactly $-2$ and $4$ .

Answer

Domain: all real $x$ except $-2,4$ ; hole $(-2,1)$ ; vertical asymptote $x=4$ .

Key idea

For a piecewise rational rule, each factor restriction must be considered within the interval where its formula applies.
Problem 7: Mark complete
Problem 8 The restricted pair

For which real constants $k$ does $|2x-1|=k$ have exactly two real solutions, both strictly between $0$ and $3$ ? Justify the restrictions on $k$ .
1. Hint 1
  How many inputs an isolated absolute value can match turns on the sign of the number it is set equal to.
2. Hint 2
  A positive target gives two candidate inputs; require each to lie inside the specified interval.
Answer

$0<k<1$ .

Full solution

Two distinct solutions require $k>0$ , since a zero right side gives one input and a negative one gives none.
For $k>0$ the inputs are
$x=\frac{1-k}{2}$
and
$x=\frac{1+k}{2}$

The smaller input is positive exactly when $k<1$ .
The larger is less than $3$ when $k<5$ , already guaranteed by $k<1$ .
Therefore $0<k<1$ .

At $k=0$ the two inputs coincide, and at $k=1$ the smaller is the excluded endpoint $0$ .
Every value strictly between them gives two allowed distinct inputs.

Answer

$0<k<1$ .

Key idea

A positive absolute-value target gives two inputs, and location requirements may further restrict that target.
Problem 8: Mark complete
Problem 9 The coefficient choice

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For $R_k(x)=\frac{kx+5}{x^2+9}$ , find its horizontal asymptote and every real $k$ for which the graph does not meet that asymptote. Justify your answer.
1. Hint 1
  Compare the degree of the top with the degree of the bottom, for each possible value of $k$ .
2. Hint 2
  Set the rule equal to the height you found, clear the denominator, and ask when the equation left in $x$ has a real solution.
Answer

Horizontal asymptote $y=0$ ; the graph misses it exactly when $k=0$ .

Full solution

The denominator $x^2+9$ is positive at every real input, so every real number is in the domain.
The numerator has degree $1$ when $k\ne0$ and degree $0$ when $k=0$ , so the bottom degree $2$ is the larger in both cases, and the horizontal asymptote is
$y=0.$

Meeting that asymptote means an output of $0$ , and a fraction is zero exactly when its numerator is, so the requirement is
$kx+5=0.$

If $k\ne0$ , that equation has the real solution $x=-5/k$ , an allowed input, so the graph meets the asymptote there.
If $k=0$ , the numerator is the constant $5$ , which is never zero, so no input works.
Hence $k=0$ is the only such value.

Answer

Horizontal asymptote $y=0$ ; the graph misses it exactly when $k=0$ .

Key idea

A rational graph reaches the height $y=0$ exactly at a numerator zero its domain still allows, so a parameter can decide whether one exists.
Problem 9: Mark complete
Problem 10 The boundary exchange

Function $f$ uses $x+1$ for $x<2$ and $x^2-1$ for $x\ge2$ . Function $g$ uses $x+1$ for $x\le2$ and $x^2-1$ for $x>2$ . Are $f$ and $g$ identical? Now let $F$ use $x+1$ for $x<1$ and $x^2-1$ for $x\ge1$ , and let $G$ use $x+1$ for $x\le1$ and $x^2-1$ for $x>1$ . Are $F$ and $G$ identical? Explain both decisions.
1. Hint 1
  The ownership exchange changes which formula is used only at the boundary itself.
2. Hint 2
  Compare both formula values at each proposed switching input.
Answer

At the boundary $2$ : $f$ and $g$ are identical. At the boundary $1$ : $F$ and $G$ are not, since $F(1)=0$ and $G(1)=2$ .

Full solution

Away from the switching input, the two functions use the same formula.
At the boundary $2$ , the formulas give
$2+1=3$
and $2^2-1=3$ , so exchanging ownership changes no value, and $f$ and $g$ are identical.

At the boundary $1$ , the two formulas give $1+1=2$ and $1^2-1=0$ .
With the stated ownership, $F(1)=0$ while $G(1)=2$ , so $F$ and $G$ differ at that input.

Answer

At the boundary $2$ : $f$ and $g$ are identical. At the boundary $1$ : $F$ and $G$ are not, since $F(1)=0$ and $G(1)=2$ .

Key idea

Exchanging boundary ownership preserves a piecewise function exactly when both formulas are defined at that boundary and agree there.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra I. You can skip it. → Continue to Chapter 15: Sequences and Series →

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