Algebra 3 Quiz: One To One And Invertibility

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One To One And InvertibilityQuestion 1 of 20

The function $f$ is defined by $f(x)=x^2-6x+10$ with domain $x\ge3$ . If $f^{-1}$ is the inverse of $f$ , which expression and domain are correct?

f^{-1}(x)=3+\sqrt{x-1}

, domain

x\ge1

f^{-1}(x)=3-\sqrt{x-1}

, domain

x\ge1

f^{-1}(x)=3+\sqrt{x+1}

, domain

x\ge -1

f^{-1}(x)=3+\sqrt{x-1}

, domain

x\ge3

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Algebra 3 Quiz

Algebra 3 Quiz: One To One And Invertibility

Practice One To One And Invertibility 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 One To One And Invertibility, 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

The function $f$ is defined by $f(x)=x^2-6x+10$ with domain $x\ge3$ . If $f^{-1}$ is the inverse of $f$ , which expression and domain are correct?

$f^{-1}(x)=3+\sqrt{x-1}$ , domain $x\ge1$ (correct answer)
$f^{-1}(x)=3-\sqrt{x-1}$ , domain $x\ge1$
$f^{-1}(x)=3+\sqrt{x+1}$ , domain $x\ge -1$
$f^{-1}(x)=3+\sqrt{x-1}$ , domain $x\ge3$

Explanation: When you see a question about inverse functions, your first move is to remember what an inverse does: it undoes the original function, swapping inputs and outputs. Since the domain of

f

x\ge3

, its range becomes the domain of

f^{-1}

. Complete the square:

f(x)=(x-3)^2+1

. With

x\ge3

, the output is at least

1

, so the range of

f

y\ge1

—therefore the domain of

f^{-1}

must be

x\ge1

. To find the rule, set

y=(x-3)^2+1

, solve:

x=3\pm\sqrt{y-1}

. Because the original domain is

x\ge3

, choose the plus branch:

f^{-1}(x)=3+\sqrt{x-1}

. That matches the correct choice. The wrong answer

3-\sqrt{x-1}

is the negative branch; it would be the inverse if the domain of

f

were

x\le3

, not

x\ge3

. The choice with

3+\sqrt{x+1}

and domain

x\ge -1

comes from misreading the completing-the-square step: if you thought you needed

x+1

, you likely shifted the constant term in the wrong direction. The choice with

3+\sqrt{x-1}

but domain

x\ge3

confuses the range of

f

with the domain of

f

—the inverse's domain should be the original function's range, not its domain. No fifth option appears here, but the same logic handles any similar distractor.estrength A quick habit: whenever you find an inverse of a quadratic, always check two things—which branch the original domain forces, and whether you used the original function's range as the inverse's domain. That one check eliminates most common traps on inverses.

Question 2

Let $f(x)=\frac{3x+2}{2x-5}$ with domain $x\ne\frac{5}{2}$ . Which statement about $f$ is true?

$f^{-1}(x)=\frac{5x+2}{2x-3}$ , domain $x\ne\frac{3}{2}$ (correct answer)
$f^{-1}(x)=\frac{5x-2}{2x-3}$ , domain $x\ne\frac{3}{2}$
$f^{-1}(x)=\frac{5x+2}{2x+3}$ , domain $x\ne-\frac{3}{2}$
$f$ does not have an inverse because its graph has a vertical asymptote at $x=\frac{5}{2}$ and a horizontal asymptote at $y=\frac{3}{2}$ .

Explanation: Whenever you need the inverse of a rational function like this, don't be thrown off by the asymptotes. The same process works as for any one-to-one function: write

y=\frac{3x+2}{2x-5}

, solve for

x

, then swap

x

and

y

. Multiplying out gives

y(2x-5)=3x+2

, so

2xy-5y=3x+2

. Collect the

x

-terms:

2xy-3x=5y+2

, or

x(2y-3)=5y+2

. Thus

x=\frac{5y+2}{2y-3}

. Swapping variables, the inverse is

\frac{5x+2}{2x-3}

,and since the denominator cannot be zero, the domain is

x\ne\frac{3}{2}

.The statement that reads exactly

\frac{5x+2}{2x-3}

with domain

x\ne\frac{3}{2}

is the true one. The statement with numerator

5x-2

has the wrong sign on the constant: solving produced

+2

, not

-2

.The statement with denominator

2x+3

and domain

x\ne-\frac{3}{2}

got the denominator sign wrong: the factor must be

2x-3

, so the excluded value is

3/2

, not

-3/2

.The statement claiming

f

has no inverse because of a vertical/horizontal asymptote confuses asymptotes with one-to-one behavior. The graph still passes the horizontal-line test on its domain; asymptotes simply identify values excluded from the domain and range. In fact, the inverse's domain exclusion

x\ne3/2

mirrors the original's horizontal asymptote

y=3/2

. Study tip: when finding inverses of rational functions, solve methodically and double-check every sign when collecting

x

-terms. Then set the denominator equal to zero to get the inverse's domain.

Question 3

Let $f(x)=x^4-2x^2+1$ . On which of the following domains is $f$ one-to-one?

$[-1,1]$
$[0,\infty)$
$(-\infty,-1]$ (correct answer)
$(-\infty,0]$

Explanation: Whenever a question asks where a function is one-to-one, imagine drawing horizontal lines: each must intersect the graph at most once on that domain. Here

f(x)=x^4-2x^2+1=(x^2-1)^2

, an even function symmetric about the

y

-axis, so intervals that contain both

x

and

-x

will generally fail. On

(-\infty,-1]

f

is strictly decreasing. Its derivative

f'(x)=4x^3-4x=4x(x-1)(x+1)

is negative for every

x<-1

, so as

x

moves from left to right,

f

falls from

\infty

0

without repeating a value. Thus this domain is one-to-one.

[-1,1]

fails because it is symmetric about

0

; for instance,

f(-1)=f(1)=0

[0,\infty)

fails because the graph decreases from

(0,1)

(1,0)

, then increases, so the horizontal line

y=\frac14

hits both

x=\sqrt{\frac12}

and

x=\sqrt{\frac32}

(-\infty,0]

fails for the mirror reason:

y=\frac14

hits

-\sqrt{\frac12}

and

-\sqrt{\frac32}

. Each wrong domain includes a turning point, causing the function to repeat output values; the trap is assuming a function is one-to-one just because it passes the vertical line test. Study tip: factor the function, then sketch or use the derivative to identify intervals where it is strictly monotonic. Only those intervals can be one-to-one.

Question 4

Which function has an inverse whose domain is $x>2$ ?

$f(x)=\log_2(x-2)$
$f(x)=2^x-2$
$f(x)=2^x+2$ (correct answer)
$f(x)=\log_2(x+2)$

Explanation: Whenever you see a question about the domain of an inverse function, translate it immediately: the domain of the inverse is exactly the range of the original function. So you need to find which function has range

x>2

(more precisely,

y>2

). The function

f(x)=2^x+2

is correct because

2^x

is always positive, so every output is greater than

2

. Its range is

(2,\infty)

, meaning the inverse has domain

x>2

. Now look at the wrong choices. The function

f(x)=\log_2(x-2)

has domain

x>2

, but its range is all real numbers, so its inverse has domain all real numbers — this is a classic trap where you confuse the original domain with the inverse's domain. The function

f(x)=2^x-2

has range

(-2,\infty)

, so its inverse's domain is

x>-2

, not

x>2

. Finally,

f(x)=\log_2(x+2)

has domain

x>-2

and range all real numbers, so its inverse's domain is also all real numbers. Your study tip: when a question asks about the inverse's domain, always ask "what is the range of the original function?" Remember that exponential functions have ranges like

(0,\infty)

, and vertical shifts move that range up or down. Log functions, on the other hand, always have range all real numbers.

Question 5

Let $f$ be defined by $f(x)=\frac{x^2-4}{x-2}$ on the domain $x\neq 2$ . Which statement about $f$ is true?

$f$ is one-to-one, and $f^{-1}(x)=x+2$ with domain $x\neq 4$ .
$f$ is one-to-one, and $f^{-1}(x)=x-2$ with domain $x\neq 2$ .
$f$ is not one-to-one because it is undefined at $x=2$ , so no inverse function exists.
$f$ is one-to-one, and $f^{-1}(x)=x-2$ with domain $x\neq 4$ . (correct answer)

Explanation: Whenever you see a rational function and are asked about inverses, simplify the function first, but keep the original domain restriction in mind. Here

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

, valid only for

x\neq 2

. The range is therefore all real numbers except

4

: if

y=x+2

and

x\neq 2

, then

y\neq 4

. Since for every

y\neq 4

exactly one

x=y-2

maps to it,

f

is one-to-one, and its inverse is

f^{-1}(x)=x-2

with domain

x\neq 4

. The distractor offering

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

simply confuses

f

with

f^{-1}

; the inverse must undo adding

2

, so it must subtract

2

. The distractor that gives

f^{-1}(x)=x-2

but domain

x\neq 2

confuses the domain of

f

with the domain of the inverse; because

4

is not in the range of

f

, the inverse must exclude

4

, not

2

. The choice that says

f

is not one-to-one because it is undefined at

x=2

misses the point: a single missing point in the domain does not destroy injectivity; it simply removes the corresponding output

4

from the range, so an inverse function exists on the domain

x\neq 4

. When attacking inverse-function questions, always simplify and compute the range. The domain of the inverse is exactly the range of the original function;it has nothing to do with where the original function is undefined. Make a habit of asking "What outputs can

f

actually produce?" before choosing the inverse's domain.

Question 6

Which function is NOT one-to-one on its stated domain?

$f(x)=2x+3$ on all real numbers
$f(x)=x^2+2x+1$ on $[0,\infty)$
$f(x)=|x-2|$ on $[2,\infty)$
$f(x)=\frac{1}{x^2+1}$ on $[-1,1]$ (correct answer)

Explanation: Whenever you see a question about one-to-one functions, think "injective" and use the horizontal line test: is there any horizontal line that touches the graph more than once? If yes, the function fails one-to-one. . The function

f(x)=\frac{1}{x^2+1}

[-1,1]

is NOT one-to-one because it is even:

f(-x)=f(x). Specifically,

f(-1)=\frac12

and

f(1)=\frac12

. Thus two different inputs,

1 and1, produce same output, so a horizontal line at

y=\frac12$$ touches both points. This symmetric “hump” shape guarantees repeated y-values across the interval. . The other choices each pass the test. The linear function

2x+3

on all real numbers is strictly increasing with slope

2

; each y-value comes from exactly one x-value, so it is one-to-one. The quadratic

x^2+2x+1

equals

(x+1)^2

; although a parabola normally fails, outer restricted domain

[0,\infty)

makes it strictly increasing because

x+1\ge 1

, so no duplicate outputs occur. Similarly,

|x-2|

[2,\infty)

simplifies to

x-2

since

x-2\ge0

, giving a straight line of slope

1

that is one-to-one. . Study tip: check the domain carefully before judging a function. Even functions like squares, absolute values, and reciprocals of even expressions often fail one-to-one on symmetric intervals, but can become one-to-one if the domain is restricted to one side of they-axis (or vertex). Here the distractor quadratic and absolute value rely exactly on that kind of domain restriction, while the reciprocal remains symmetric on a symmetric interval.

Question 7

Let $f(x)=2x+3$ if $x<0$ , and $f(x)=x^2+1$ if $x\ge0$ . Which statement best describes $f$ ?

$f$ is one-to-one because each piece is one-to-one.
$f$ is not one-to-one because $f(-1)=f(0)=1$ . (correct answer)
$f$ is one-to-one because its range is all real numbers.
$f$ is not one-to-one because $x^2+1$ is not one-to-one on $[0,\infty)$ .

Explanation: Whenever you see a piecewise function and are asked whether it is one-to-one, don't just check each formula separately. A function is one-to-one only if no output value is shared by two different input values across the entire domain. Here, look at the boundary between the pieces: for

x<0

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

, and for

x\ge0

f(0)=0^2+1=1

. Since

-1\ne0

but

f(-1)=f(0)

, the function repeats an output, so it is not one-to-one. The choice saying "not one-to-one because

f(-1)=f(0)=1

" gives exactly that evidence and is correct. The choice saying "one-to-one because each piece is one-to-one" misses the point: each piece can be one-to-one on its own, but the pieces can still overlap in output values across the boundary. The choice saying "one-to-one because its range is all real numbers" confuses being onto with being one-to-one; having range all real numbers means the function hits every real number, not that every output comes from only one input. Finally, the choice saying "not one-to-one because

x^2+1

is not one-to-one on

[0,\infty)

" has the right conclusion but the wrong reason: on

[0,\infty)

x^2+1

is actually increasing and therefore one-to-one. The real failure is the overlap between the two pieces. For piecewise one-to-one questions, evaluate each formula at boundary points and compare outputs. A single duplicated

y

-value is enough to disprove one-to-one.

Question 8

Let $f(x)=x^2+1$ with domain $x\ge0$ , and let $g(x)=\sqrt{x-1}$ with domain $x\ge1$ . Which statement correctly verifies that $g$ is the inverse of $f$ ?

$f$ and $g$ are inverses because $f(0)=1$ and $g(1)=0$ .
$f(g(x))=x$ for all real $x$ .
$g(f(x))=x$ for all real $x$ .
$f(g(x))=x$ for $x\ge1$ and $g(f(x))=x$ for $x\ge0$ . (correct answer)

Explanation: When you see an inverse-function question, remember: two functions are inverses only if their compositions return

x

, and only on the correct domains. Start by testing both compositions. Since

g(x)=\sqrt{x-1}

, we have

f(g(x))=(\sqrt{x-1})^2+1=x-1+1=x.

But

g(x)

is only defined for

x\ge1

, so this holds for

x\ge1

, not for all real

x

. Next,

g(f(x))=\sqrt{(x^2+1)-1}=\sqrt{x^2}=|x|.

Because

f

's domain is

x\ge0

|x|=x

here, so

g(f(x))=x

for

x\ge0

. Therefore, the statement "

f(g(x))=x

for

x\ge1

and

g(f(x))=x

for

x\ge0

" correctly verifies that

g

is the inverse of

f

. Why are the others wrong? The statement using only

f(0)=1

and

g(1)=0

checks just one point, not every input. The claim that

f(g(x))=x

for all real

x

ignores that

g

is undefined for

x<1

. The claim that

g(f(x))=x

for all real

x

fails for negative

x

, because

\sqrt{x^2}=|x|

, not

x

. Study tip: to verify inverses, always check both compositions and respect the given domains. One matching pair of values is never enough.

Question 9

A student claims that $f(x)=x^3-x$ is one-to-one on all real numbers because every cubic polynomial has opposite end behavior. Which statement correctly evaluates the claim?

The claim is correct; opposite end behavior guarantees that each horizontal line intersects the graph exactly once.
The claim is incorrect; for example, $f(-1)=f(1)=0$ , so two different inputs have the same output. (correct answer)
The claim is incorrect; an odd-degree polynomial can never be one-to-one.
The claim is correct; $x^3$ and $x$ are both one-to-one, so their difference is one-to-one.

Explanation: Whenever you see a question about whether a function is one-to-one, the core definition is key: no two different inputs can share the same output, and the horizontal line test is your visual check. The student's reasoning about opposite end behavior is a classic trap because end behavior only describes what happens far to the left and right; it says nothing about the middle of the graph. To evaluate the claim, test specific values. For

f(x)=x^3-x

, notice that

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

and

f(1)=1^3-1 = 0

. Since

f(-1)=f(1)=0

, two different inputs,

-1

and

1

, produce the same output, directly violating the definition of one-to-one. The claim is therefore incorrect. Now, examine the wrong choices. The statement that "opposite end behavior guarantees that each horizontal line intersects the graph exactly once" is false—end behavior only concerns the tails, and a cubic can have a local maximum and minimum, causing it to double back and fail the horizontal line test. The statement that "an odd-degree polynomial can never be one-to-one" is also false; for example,

f(x)=x^3

is an odd-degree polynomial that is strictly increasing and therefore one-to-one. Finally, the statement that "

x^3

and

x

are both one-to-one, so their difference is one-to-one" is a flawed assumption—the difference of two one-to-one functions is not guaranteed to be one-to-one. Your takeaway: when checking for one-to-one, always plug in symmetric values like

1

and

-1

, and use the horizontal line test rather than relying on end behavior or properties of sums and differences.

Question 10

Let $f(x)=2^{x-3}-5$ . What are the domain and range of $f^{-1}$ ?

domain $x>-5$ , range all real numbers (correct answer)
domain $x\ge -5$ , range all real numbers
domain all real numbers, range $y>-5$
domain $x>5$ , range all real numbers

Explanation: Whenever you're asked about the domain and range of an inverse function, remember that the inverse swaps the roles of inputs and outputs. So the domain of

f^{-1}

equals the range of

f

, and the range of

f^{-1}

equals the domain of

f

. Here,

f(x)=2^{x-3}-5

is an exponential function. Its domain is all real numbers because you can plug in any

x

. Its range is

y>-5

, since

2^{x-3}

is always positive, so subtracting 5 gives values strictly greater than

-5

, approaching

-5

as a horizontal asymptote but never reaching it. Therefore, for

f^{-1}

, the domain is

x>-5

(the range of

f

), and the range is all real numbers (the domain of

f

). The choice "domain

x\ge -5

, range all real numbers" is tempting but incorrect because

-5

is not actually achieved by

f

, so it cannot be in the domain of

f^{-1}

. The choice "domain all real numbers, range

y>-5

" describes the original function

f

, not its inverse — it reverses the two. "Domain

x>5

, range all real numbers" uses the wrong sign; the horizontal asymptote is at

-5

, not

5

, and there is no shift by

+5

in this function. Study tip: for inverse functions, just swap the domain and range of the original. Also, watch for open versus closed endpoints—exponential functions never actually touch their horizontal asymptote.

Question 11

The one-to-one function $f$ satisfies $f(2)=5$ , $f(5)=7$ , and $f(7)=2$ . If $g$ is the inverse of $f$ , what is $g(f(7))+g(2)$ ?

5
7
14 (correct answer)
Cannot be determined

Explanation: Whenever you see a function and its inverse, focus on reversing the input-output pairs. Since

f

gives

2\to 5

5\to 7

, and

7\to Car2

, its inverse

g

gives

5\to Car2

7\to Car5

, and

2\to Car7

. Now evaluate carefully. The inner value is

f(7)=2

. Therefore

g(f(7))=g(2)

, and from the inverse pairs

g(2)=7

. The second term is also

g(2)=7

. So [ g(f(7))+g(2)=7+7=14. The choice

5

comes from confusing

g

with

f

—for example, using

f(2)=5

to claim

g(2)=5

. But

f(2)=5

actually means

g(5)=2

, not the other way around. The choice

7

ignores the second term: it gives the value of

g(f(7))

alone, or it treats

+g(2)

as adding zero. The choice "Cannot be determined" overlooks that although not every value of

f

is given, the values needed here are fully determined by the inverted pairs. Study tip: when inverse-function values appear, first write down the reversed pairs from the given facts. That simple step stops the most common trap on this exam: using the original function's table where you needthe inverse's.

Question 12

Which condition guarantees that a function $f$ has an inverse?

The graph of $f$ passes the vertical line test.
For all $a,b$ in the domain of $f$ , if $f(a)=f(b)$ , then $a=b$ . (correct answer)
$f$ is an even function whose domain is all real numbers.
The range of $f$ equals the domain of $f$ , so the inverse has the same domain.

Explanation: Whenever you see a question about inverse functions, your first thought should be: an inverse exists only if each output comes from exactly one input. That is, the function must be one-to-one. The condition "for all

a,b

in the domain of

f

, if

f(a)=f(b)

, then

a=b

" is the precise definition of one-to-one. It guarantees that when you reverse the function, there is no ambiguity about which input produced a given output. The vertical line test only tells you that a graph represents a function, not that the function has an inverse. An even function with domain all real numbers, such as

f(x)=x^2

, actually fails the one-to-one condition because

f(1)=f(-1)

, so it cannot have an inverse. And just having the range equal the domain says nothing about whether different inputs could collide; for example, a constant function on a one-element domain technically works, but on a larger domain it fails because many inputs share the same output. So when you see inverse-function questions, ask yourself: Could two different inputs produce the same output? If yes, no inverse. Graphically, check the horizontal line test — any horizontal line should hit the curve at most once. That visual habit will help you quickly eliminate tempting but incomplete conditions.

Question 13

For which real value(s) of $k$ does $f(x)=x^2+2x+k$ , restricted to the domain $x\le -1$ , have an inverse function?

All real $k$ , because changing $k$ shifts the graph vertically and does not change one-to-one behavior. (correct answer)
Only $k=1$ , because the vertex must lie on the $x$ -axis.
Only $k\ge 1$ , so that $f(x)\ge 0$ on the restricted domain.
No real $k$ , because a quadratic is symmetric about its axis and fails the horizontal line test on any interval.

Explanation: The vertex of

x^2+2x+k

is at

x=-1

. On

x\le -1

, the function is strictly decreasing, so every horizontal line meets the graph at most once. The constant

k

only shifts the graph vertically; it changes the range but does not affect whether the function is one-to-one. Therefore every real

k

works. The other choices confuse the location or range of the vertex with injectivity.

Question 14

Let $f(x)=\frac{x^2-4}{x-2}$ , with domain all real numbers except 2. Which statement is true?

$f$ is not one-to-one because $x^2-4$ has two $x$ -intercepts.
$f$ is one-to-one and $f^{-1}(x)=x-2$ for all real $x$ .
$f$ is one-to-one and $f^{-1}(x)=x-2$ , with domain $x\ne 4$ . (correct answer)
$f$ is one-to-one and $f^{-1}(x)=x+2$ , with domain $x\ne 2$ .

Explanation: For

x\ne2

, the expression simplifies to

x+2

, so

f

is injective: equal outputs force equal inputs. Its range is all real numbers except 4, because

f(2)

is undefined. Therefore the inverse is

f^{-1}(x)=x-2

with domain

x\ne4

. Saying the inverse is defined for all real

x

ignores the missing output 4, and

x+2

is not the inverse.

Question 15

For what real value of $a$ is $f(x)=\frac{ax+1}{x-2}$ its own inverse? Assume the natural domain $x\ne2$ .

$a=-2$
$a=1$
$a=2$ (correct answer)
$a=5$

Explanation: Solving

y=\frac{ax+1}{x-2}

for

x

gives

x=\frac{2y+1}{y-a}

, so

f^{-1}(x)=\frac{2x+1}{x-a}

. Setting

f^{-1}(x)=f(x)

gives

\frac{ax+1}{x-2}=\frac{2x+1}{x-a}

. Cross-multiplying and equating coefficients yields

a=2

. The other values do not make the two formulas identical.

Question 16

Which function is one-to-one and has an inverse whose domain is all real numbers except 0?

$f(x)=\frac{x+2}{x+1}$ , $x\ne -1$
$f(x)=\frac{x}{x^2-1}$ , $x\ne \pm1$
$f(x)=x^2+1$
$f(x)=\frac{1}{x+1}$ , $x\ne -1$ (correct answer)

Explanation: The inverse of a one-to-one function has domain equal to the range of the original function. For

f(x)=\frac{1}{x+1}

, the numerator is never 0, so the range is all real numbers except 0. The function is also injective. The first rational function has range all real numbers except 1, the second is not one-to-one, and the quadratic is not one-to-one on all real numbers.

Question 17

Suppose $f(g(x))=x$ for every $x$ in the domain of $g$ , but $g(f(x))\ne x$ for some $x$ in the domain of $f$ . Which statement must be true?

$f$ is one-to-one.
$f$ and $g$ are inverse functions.
$g$ is the reciprocal of $f$ .
$g$ is one-to-one. (correct answer)

Explanation: If

g(a)=g(b)

, then applying

f

gives

a=f(g(a))=f(g(b))=b

, so

g

must be one-to-one. One composition alone is not enough to make two functions inverses, so

f

need not be one-to-one and the pair need not be inverse functions. Reciprocal has no connection to inverse functions.

Question 18

The function $f(x)=x^5+x^3+x$ is one-to-one on $\mathbb{R}$ . If $g=f^{-1}$ , what is $g(3)$ ?

1 (correct answer)
3
27
273

Explanation: By definition,

g(3)

is the input whose output is 3. Since

f(1)=1^5+1^3+1=3

, the inverse value is

g(3)=1

. The value 273 is

f(3)

, not

g(3)

, and 3 is not generally the inverse of 3.

Question 19

Which function is one-to-one on all real numbers?

$f(x)=x^2-2x+1$
$f(x)=|x+1|+|x-1|$
$f(x)=\frac{x^3}{x^2+1}$ (correct answer)
$f(x)=\cos x$

Explanation: The quadratic factors as

(x-1)^2

, so it has symmetric outputs. The absolute-value expression is constant on

[-1,1]

, and cosine is periodic. For

f(x)=\frac{x^3}{x^2+1}

, if

f(a)=f(b)

, then

(a-b)(a^2b^2+a^2+ab+b^2)=0

. The second factor is positive when

a\ne b

, so

a=b

. Thus it is one-to-one.

Question 20

The function $f(x)=2x^2-8x+5$ is restricted to one side of its vertex so that it is one-to-one. Which restriction makes the range of $f^{-1}$ equal to $[2,\infty)$ ?

$(-\infty,2]$
$[2,\infty)$ (correct answer)
$[0,\infty)$
$(-\infty,0]$

Explanation: The range of an inverse function is the domain of the original function. The vertex is at

x=2

, and the right-side restriction

[2,\infty)

makes

f

strictly increasing. The left-side restriction

(-\infty,2]

would make the inverse range

(-\infty,2]

. The interval

[0,\infty)

is not one-sided about the vertex and is not one-to-one.