Math 3 Quiz: One To One Functions

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One To One FunctionsQuestion 1 of 16

The function $h(x) = \frac{ax + b}{cx + d}$ where $ad - bc \neq 0$ is defined on $\mathbb{R} \setminus \{-\frac{d}{c}\}$ . For which condition on the parameters is $h$ guaranteed to be one-to-one on its domain?

ad - bc > 0

, ensuring the function is strictly increasing on each connected component

c \neq 0

, which guarantees the function has exactly one vertical asymptote and no horizontal asymptote

ad - bc \neq 0

, which is the condition already given and sufficient for injectivity

a \neq 0

and

c \neq 0

, preventing the function from reducing to a constant or linear form

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Math 3 Quiz

Math 3 Quiz: One To One Functions

Practice One To One Functions in Math 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 Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 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

$ad - bc > 0$ , ensuring the function is strictly increasing on each connected component
$c \neq 0$ , which guarantees the function has exactly one vertical asymptote and no horizontal asymptote
$ad - bc \neq 0$ , which is the condition already given and sufficient for injectivity (correct answer)
$a \neq 0$ and $c \neq 0$ , preventing the function from reducing to a constant or linear form

Explanation: A rational function of the form

h(x) = \frac{ax + b}{cx + d}

(a Möbius transformation) is one-to-one on its domain if and only if

ad - bc \neq 0

. This condition ensures the function is invertible. To verify: if

h(x_1) = h(x_2)

, then

\frac{ax_1 + b}{cx_1 + d} = \frac{ax_2 + b}{cx_2 + d}

. Cross-multiplying and simplifying leads to

(ad - bc)(x_1 - x_2) = 0

. If

ad - bc \neq 0

, then

x_1 = x_2

. Choice A is incorrect because the sign of

ad - bc

doesn't matter for injectivity. Choice B is wrong because

c \neq 0

alone isn't sufficient. Choice D is incorrect because the function can be one-to-one even when

c = 0

(making it linear) as long as

a \neq 0

Question 2

Consider the function $f(x) = x^3 - 3x^2 + 2x$ on the interval $[0, 3]$ . To determine if $f$ is one-to-one on this domain, which of the following approaches would provide the most definitive conclusion?

Check that $f'(x) > 0$ for all $x$ in $[0, 3]$ , since positive derivative guarantees one-to-one behavior
Verify that $f$ has no repeated values by solving $f'(x) = 0$ and checking function behavior at critical points (correct answer)
Calculate $f(0)$ , $f(1.5)$ , and $f(3)$ to confirm the function is strictly increasing throughout the interval
Apply the horizontal line test conceptually since $f$ is a cubic polynomial with positive leading coefficient

Explanation: To determine if a function is one-to-one, we need to check if it ever has the same output for different inputs. For

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

, we find

f'(x) = 3x^2 - 6x + 2

. Setting

f'(x) = 0

gives critical points, and we must check if the function has local maxima and minima that could create repeated values. Choice A is incorrect because

f'(x)

is not always positive on

[0,3]

. Choice C only checks three points, which is insufficient. Choice D is incorrect because the leading coefficient alone doesn't determine one-to-one behavior on a restricted domain.

Question 3

A student claims that the function $g(x) = \frac{x}{x^2 + 1}$ is one-to-one on the interval $[-1, 1]$ because it passes through the origin and appears to be increasing near $x = 0$ . Which analysis most accurately evaluates this claim?

The claim is correct because $g'(0) = 1 > 0$ , indicating the function is increasing at the origin
The claim is incorrect because $g'(x) = \frac{1-x^2}{(x^2+1)^2}$ changes sign within $[-1, 1]$ , but the function is still one-to-one there (correct answer)
The claim is incorrect because the function has a horizontal asymptote, which means it must repeat values
The claim is correct because the denominator $x^2 + 1$ is always positive, preventing undefined behavior

Explanation: To determine if

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

is one-to-one on

[-1, 1]

, we find

g'(x) = \frac{1-x^2}{(x^2+1)^2}

. This derivative is positive for

|x| < 1

and equals zero at

x = \pm 1

. Since

g'(x) > 0

throughout

(-1, 1)

, the function is strictly increasing on

[-1, 1]

, making it one-to-one despite the student's flawed reasoning. Choice A gives correct conclusion but insufficient reasoning. Choice C is irrelevant to the given interval. Choice D addresses continuity, not one-to-one property. The student's reasoning about behavior 'near zero' is insufficient for the entire interval.

Question 4

A function $h(x)$ satisfies the property that $h(a) = h(b)$ implies $a = b$ for all $a, b$ in its domain $D$ . If $D = [-3, 0) \cup (0, 3]$ , which statement about $h(x)$ is necessarily true?

$h(x)$ is one-to-one on $D$ , and this property is unaffected by the discontinuity at $x = 0$ (correct answer)
$h(x)$ cannot be one-to-one because the domain is not connected, creating potential for repeated values
$h(x)$ is one-to-one on each piece of the domain separately, but not on the entire domain $D$
The one-to-one property depends on the specific function values at the boundary points $x = -3$ and $x = 3$

Explanation: The given property '

h(a) = h(b)

implies

a = b

for all

a, b

in domain

D

' is precisely the definition of a one-to-one function. This property holds regardless of whether the domain is connected or has discontinuities. The fact that

x = 0

is not in the domain doesn't affect the one-to-one property, as this property only concerns pairs of points that are both in the domain. Choice B incorrectly assumes disconnected domains prevent one-to-one functions. Choice C contradicts the given condition. Choice D incorrectly focuses on boundary values rather than the given logical condition.

Question 5

A function $h(x)$ is defined piecewise as $h(x) = \begin{cases} x^2 - 4 & \text{if } x \leq 0 \\ -x + 4 & \text{if } x > 0 \end{cases}$ . For $h(x)$ to be one-to-one on its entire domain, which modification would be necessary?

Restrict the first piece to $x \leq -2$ to eliminate the decreasing portion of the parabola
Change the second piece to $x + 4$ to make both pieces increasing functions
Restrict the domain to $x \geq 0$ to use only the linear portion of the function (correct answer)
No modification needed since the function is already one-to-one as defined

Explanation: For the current piecewise function, the first piece

x^2 - 4

x \leq 0

gives values from

-4

+\infty

, while the second piece

-x + 4

x > 0

gives values from

(-\infty, 4)

. These ranges overlap significantly, meaning the same y-value can come from both pieces, violating the one-to-one property. Choice C restricts to only the linear piece, which is strictly decreasing and therefore one-to-one. Choice A doesn't solve the overlap problem. Choice B creates two increasing pieces that still overlap. Choice D is incorrect because the function has repeated y-values.

Question 6

Consider the function $f(x) = x^3 + ax^2 + bx + c$ where $a$ , $b$ , and $c$ are constants. For this function to be one-to-one on its entire domain $(-\infty, \infty)$ , which condition on the discriminant of $f'(x)$ must be satisfied?

The discriminant $4a^2 - 12b$ must be positive to ensure $f'(x)$ has two distinct real roots
The discriminant condition is irrelevant since all cubic functions with positive leading coefficient are one-to-one
The discriminant $4a^2 - 12b$ must equal zero to ensure $f'(x)$ has exactly one repeated root
The discriminant $4a^2 - 12b$ must be negative to ensure $f'(x)$ has no real roots (correct answer)

Explanation: When you encounter questions about one-to-one functions, you need to think about when a function passes the horizontal line test—meaning no horizontal line intersects the graph more than once. For polynomial functions, this connects directly to the behavior of the derivative. A cubic function

f(x) = x^3 + ax^2 + bx + c

is one-to-one when it's always increasing (since the leading coefficient is positive). This happens when

f'(x) \geq 0

for all

x

, and more specifically, when

f'(x) > 0

everywhere except possibly at isolated points. The derivative is

f'(x) = 3x^2 + 2ax + b

, which is a quadratic with positive leading coefficient. For

f'(x)

to never become negative, it cannot have two distinct real roots (which would create a valley where

f'(x) < 0

). The discriminant of this quadratic is

(2a)^2 - 4(3)(b) = 4a^2 - 12b

. When this discriminant is negative,

f'(x)

has no real roots and stays positive everywhere, ensuring

f(x)

is strictly increasing and therefore one-to-one. Option A is wrong because two distinct roots would create intervals where

f'(x) < 0

, making the function decrease. Option B incorrectly assumes all cubics with positive leading coefficients are one-to-one—this ignores the critical role of the middle terms. Option C is wrong because a repeated root would create a point where

f'(x) = 0

, potentially allowing the function to have a horizontal tangent and fail the one-to-one test. Remember: for polynomials to be one-to-one, focus on when the derivative maintains its sign. The discriminant tells you about the roots of that derivative.

Question 7

Consider the function $f(x) = |x - 2| + |x + 1|$ on different domains. Which statement about the one-to-one property of this function is correct?

$f(x)$ is one-to-one on $(-\infty, -1]$ because it's decreasing, and on $[2, \infty)$ because it's increasing
$f(x)$ is one-to-one only on $[2, \infty)$ because this is where both absolute value expressions are positive
$f(x)$ is one-to-one on $(-\infty, -1]$ and $[2, \infty)$ , but not on any interval containing $[-1, 2]$ (correct answer)
$f(x)$ is never one-to-one on any interval because absolute value functions always fail the horizontal line test

Explanation: To analyze

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

, we consider the critical points

x = -1

and

x = 2

. For

x \leq -1

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

(decreasing). For

-1 < x < 2

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

(constant). For

x \geq 2

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

(increasing). The function is one-to-one on

(-\infty, -1]

and

[2, \infty)

where it's strictly monotonic, but not on any interval containing part of

[-1, 2]

where it's constant. Choice A incorrectly states the monotonicity. Choice B misses the decreasing portion. Choice D is false about absolute value functions in general.

Question 8

Two students are debating whether $g(x) = \ln(x^2 + 1)$ is one-to-one on $\mathbb{R}$ . Student A argues it's not one-to-one because $x^2 + 1$ is even. Student B argues it is one-to-one because the natural logarithm is always one-to-one. Which analysis is most mathematically sound?

Student A is correct; since $g(-x) = g(x)$ for all $x$ , the function is even and cannot be one-to-one (correct answer)
Student B is correct; the composition of one-to-one functions is always one-to-one, so $g(x)$ inherits this property
Both students are partially correct; $g(x)$ is one-to-one on $(0, \infty)$ but not on all of $\mathbb{R}$
Neither student's reasoning is complete; the correct approach requires checking if $g(x_1) = g(x_2)$ implies $x_1 = x_2$

Explanation: Student A correctly identifies that

g(x) = \ln(x^2 + 1)

is an even function because

g(-x) = \ln((-x)^2 + 1) = \ln(x^2 + 1) = g(x)

. Any even function that is not constant cannot be one-to-one on

\mathbb{R}

because it produces the same output for

x

and

-x

when

x \neq 0

. Student B's reasoning is flawed because while

\ln(u)

is one-to-one,

u = x^2 + 1

is not one-to-one as a function of

x

. Choice B incorrectly applies composition properties. Choice C is partially correct about intervals but doesn't address the students' reasoning. Choice D suggests an unnecessary approach when symmetry arguments suffice.

Question 9

Consider $f(x) = e^{x^2 - 2x}$ . To find the maximal intervals on which $f$ is one-to-one, which approach yields the correct answer?

Find where $f'(x) = 0$ : this gives $x = 1$ , so $f$ is one-to-one on $(-\infty, 1]$ and $[1, \infty)$ separately (correct answer)
Since $e^u$ is always one-to-one, $f$ is one-to-one on $\mathbb{R}$ regardless of the exponent $u$
Since $x^2 - 2x = (x-1)^2 - 1 \geq -1$ , the function $f$ has minimum value $e^{-1}$ and is therefore one-to-one on $\mathbb{R}$
Analyze where $x^2 - 2x$ is monotonic: it decreases on $(-\infty, 1]$ and increases on $[1, \infty)$ but this doesn't determine one-to-one intervals

Explanation: When determining where a function is one-to-one, you need to find intervals where the function is strictly monotonic (either always increasing or always decreasing). The key insight is that a function can only be one-to-one on an interval if it never "turns around" – that is, if its derivative doesn't change sign. The correct approach is to find where

f'(x) = 0

to locate critical points. For

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

, using the chain rule:

f'(x) = e^{x^2 - 2x} \cdot (2x - 2) = e^{x^2 - 2x} \cdot 2(x - 1)

. Since

e^{x^2 - 2x} > 0

always,

f'(x) = 0

only when

x = 1

. For

x < 1

f'(x) < 0

(decreasing), and for

x > 1

f'(x) > 0

(increasing). Therefore,

f

is one-to-one on

(-\infty, 1]

and

[1, \infty)

separately, making A correct. Option B is wrong because while

e^u

is one-to-one as a function of

u

, when

u = x^2 - 2x

is not one-to-one in

x

, the composition

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

won't be one-to-one either. Option C incorrectly assumes that having a minimum value makes a function one-to-one – this confuses range restrictions with injectivity. Option D correctly identifies the monotonic intervals but wrongly suggests this doesn't determine the answer. Remember: To find maximal one-to-one intervals, always locate critical points using the derivative, then determine where the function is strictly monotonic between these points.

Question 10

A function $g$ is defined such that $g(2x - 1) = x^3 + 1$ for all real numbers $x$ . To determine if $g$ is one-to-one on its natural domain, which approach is most mathematically sound?

Show that $g'(x) > 0$ for all $x$ in the domain after finding $g(x)$ explicitly
Verify that $x^3 + 1$ is one-to-one, since composition preserves one-to-one properties
Check if $2x - 1$ is one-to-one, then apply the composition rule for injective functions
Assume $g(a) = g(b)$ and show $a = b$ using the given functional equation directly (correct answer)

Explanation: The most direct approach is to use the definition of one-to-one. If

g(a) = g(b)

, we need to show

a = b

. From the given equation, if

a

and

b

are in the range of

2x-1

(which is all real numbers), then

a = 2x_1 - 1

and

b = 2x_2 - 1

for some

x_1, x_2

. Then

g(a) = x_1^3 + 1

and

g(b) = x_2^3 + 1

. If these are equal, then

x_1^3 = x_2^3

, so

x_1 = x_2

, which means

a = b

. Choice A requires finding

g(x)

explicitly, which is unnecessary. Choice B incorrectly assumes composition preserves injectivity in this direction. Choice C misunderstands the relationship between the functions.

Question 11

A function $f: \mathbb{R} \to \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y) + 2xy$ for all $x, y \in \mathbb{R}$ . If $f(1) = 3$ , determine whether $f$ can be one-to-one.

Yes, because the functional equation uniquely determines $f(x) = x^2 + 2x$ , which is one-to-one on $[-1, \infty)$
No, because any function satisfying this functional equation must have $f(0) = 0$ and be even
Yes, because setting $y = -x$ in the equation shows $f$ is strictly increasing everywhere
No, because the functional equation forces $f(x) = x^2 + 2x$ , and quadratics are not globally one-to-one (correct answer)

Explanation: Let's find

f

explicitly. Setting

y = 0

f(x + 0) = f(x) + f(0) + 2x \cdot 0

, so

f(x) = f(x) + f(0)

, which gives

f(0) = 0

. Setting

x = y = 1

f(2) = f(1) + f(1) + 2 = 6 + 2 = 8

. Setting

x = 2, y = -1

f(1) = f(2) + f(-1) + 2(2)(-1) = 8 + f(-1) - 4

, so

3 = 4 + f(-1)

, giving

f(-1) = -1

. By induction and the functional equation's structure, we can show

f(x) = x^2 + 2x

. Verification:

f(x+y) = (x+y)^2 + 2(x+y) = x^2 + 2xy + y^2 + 2x + 2y = (x^2 + 2x) + (y^2 + 2y) + 2xy = f(x) + f(y) + 2xy

✓. Since

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

, we have

f'(x) = 2x + 2 = 0

when

x = -1

. The function decreases on

(-\infty, -1)

and increases on

(-1, \infty)

, so it's not globally one-to-one. For example,

f(-2) = 4 - 4 = 0 = f(0)

Question 12

Consider the inverse trigonometric function $g(x) = \arctan(x^2 - 1)$ . On which of the following intervals is $g$ one-to-one?

$(-\infty, \infty)$ , since $\arctan$ is one-to-one and $x^2 - 1$ maps to all real numbers
$[-1, 1]$ , where $x^2 - 1 \leq 0$ ensures $g(x) \leq 0$ and avoids branch complications
$[1, \infty)$ , where $x^2 - 1$ is strictly increasing and $\arctan$ preserves this property (correct answer)
$(-\infty, -1] \cup [1, \infty)$ , where $x^2 - 1 \geq 0$ and the function is well-defined

Explanation: When analyzing whether a composite function like

g(x) = \arctan(x^2 - 1)

is one-to-one, you need to examine both the inner function

x^2 - 1

and how the outer function

\arctan

behaves with it. Since

\arctan

is strictly increasing on all real numbers,

g(x)

will be one-to-one precisely where the inner function

x^2 - 1

is one-to-one. The parabola

x^2 - 1

decreases on

(-\infty, 0]

and increases on

[0, \infty)

, so it's only one-to-one on intervals that don't cross

x = 0

. Answer C is correct because on

[1, \infty)

, the function

x^2 - 1

is strictly increasing. Since

\arctan

is also strictly increasing, their composition

g(x)

remains strictly increasing and therefore one-to-one on this interval. Answer A is wrong because

x^2 - 1

is not one-to-one over all real numbers—it takes the same values on both sides of

x = 0

. For example,

g(-2) = g(2) = \arctan(3)

. Answer B fails because while

g(x) \leq 0

[-1, 1]

, the function still isn't one-to-one since

x^2 - 1

decreases then increases on this interval, creating the same duplicate value problem. Answer D is incorrect because although

x^2 - 1 \geq 0

(-\infty, -1] \cup [1, \infty)

, this union includes both decreasing and increasing portions of the parabola, so

g

won't be one-to-one across the entire set. Study tip: For composite functions to be one-to-one, focus on finding intervals where the inner function is monotonic (strictly increasing or decreasing).

Question 13

Consider $h(x) = \ln(x^2 - 4x + 5)$ . To determine if $h$ is one-to-one on its natural domain, which analysis is correct?

Since $x^2 - 4x + 5 > 0$ for all real $x$ , and $\ln$ is one-to-one, $h$ is one-to-one on $\mathbb{R}$
The function $h$ is one-to-one on $[2, \infty)$ where $x^2 - 4x + 5$ increases, but not on $(-\infty, \infty)$ (correct answer)
Since the discriminant $16 - 20 = -4 < 0$ , the quadratic has no real roots, making $h$ undefined
The function $h$ is one-to-one only on intervals where $x^2 - 4x + 5 > 1$ , ensuring $\ln$ output is positive

Explanation: First, note that

x^2 - 4x + 5 = (x-2)^2 + 1 \geq 1 > 0

for all real

x

, so

h

is defined on all of

\mathbb{R}

. While

\ln

is one-to-one, the composition

h(x) = \ln(g(x))

is one-to-one only if

g(x) = x^2 - 4x + 5

is one-to-one. Since

g(x)

is a parabola opening upward with vertex at

x = 2

, it decreases on

(-\infty, 2]

and increases on

[2, \infty)

. Therefore

g

is not one-to-one on

\mathbb{R}

(e.g.,

g(1) = g(3) = 2

), so

h

is not one-to-one on

\mathbb{R}

. However,

h

is one-to-one on

[2, \infty)

where

g

is strictly increasing. Choice A incorrectly assumes composition of one-to-one functions is always one-to-one. Choice C misapplies the discriminant. Choice D incorrectly focuses on the sign of

\ln

output rather than injectivity.

Question 14

Let $f(x) = \sqrt{x^2 - 6x + 8}$ and consider its natural domain. If we restrict $f$ to make it one-to-one, which of the following represents the largest possible domain for such a restriction?

$(-\infty, 2] \cup [4, \infty)$ , using the complete natural domain with appropriate monotonic pieces
$[3, \infty)$ , restricting to where the expression under the radical is minimized and increasing
$(-\infty, 2]$ , using only the left branch where the function decreases monotonically
$[4, \infty)$ , using only the right branch where the function increases monotonically (correct answer)

Explanation: First, find the natural domain:

x^2 - 6x + 8 \geq 0

, which factors as

(x-2)(x-4) \geq 0

. This gives domain

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

. The expression

x^2 - 6x + 8 = (x-3)^2 - 1

has vertex at

x = 3

. On

(-\infty, 2]

, we have

x - 3 < 0

, so

(x-3)^2

decreases as

x

increases toward 2, making

f(x)

decrease. On

[4, \infty)

, we have

x - 3 > 0

, so

(x-3)^2

increases as

x

increases, making

f(x)

increase. Both intervals alone give one-to-one functions, but

[4, \infty)

is larger than

(-\infty, 2]

. Choice A is wrong because the complete domain isn't one-to-one (

f(1) = f(5) = \sqrt{3}

). Choice B uses an invalid domain since

f(3)

is undefined. Choice C gives a smaller domain than choice D.

Question 15

A function $f$ is defined implicitly by $x^3 + y^3 = 6xy$ near the point $(3, 3)$ . In a neighborhood of this point, is $y$ a one-to-one function of $x$ ?

Yes, because implicit differentiation gives $\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x} \neq 0$ near $(3,3)$ (correct answer)
No, because the curve $x^3 + y^3 = 6xy$ has multiple branches that intersect near $(3,3)$
Yes, because the implicit function theorem guarantees local existence and uniqueness of $y = f(x)$
Cannot determine without explicitly solving for $y$ in terms of $x$ from the given equation

Explanation: Using implicit differentiation on

x^3 + y^3 = 6xy

3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}

. Solving for

\frac{dy}{dx}

3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2

, so

\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}

. At

(3,3)

\frac{dy}{dx} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1 \neq 0

. By the implicit function theorem, since

\frac{\partial}{\partial y}(x^3 + y^3 - 6xy) = 3y^2 - 6x = 27 - 18 = 9 \neq 0

(3,3)

, there exists a neighborhood where

y

is a unique function of

x

. Since

\frac{dy}{dx} \neq 0

in this neighborhood (by continuity), the function is strictly monotonic and hence one-to-one locally. Choice C mentions the implicit function theorem but doesn't verify the conditions. Choice B is incorrect about intersecting branches locally. Choice D is unnecessarily restrictive.

Question 16

The function $f(x) = |x - a| + |x - b|$ where $a < b$ is defined for all real $x$ . On which interval is $f$ guaranteed to be one-to-one?

$(-\infty, a]$ , where both absolute value expressions simplify and $f$ becomes linear
$[b, \infty)$ , where the derivative exists and is constant, ensuring strict monotonicity (correct answer)
$[a, b]$ , where the function achieves its minimum value and remains constant
$(-\infty, \frac{a+b}{2}]$ , where the function decreases toward its global minimum point

Explanation: We need to analyze

f(x) = |x - a| + |x - b|

where

a < b

. For

x < a

: both

(x-a) < 0

and

(x-b) < 0

, so

f(x) = -(x-a) - (x-b) = -2x + (a+b)

, which decreases with slope

-2

. For

a \leq x \leq b

(x-a) \geq 0

and

(x-b) \leq 0

, so

f(x) = (x-a) - (x-b) = b-a

, which is constant. For

x > b

: both

(x-a) > 0

and

(x-b) > 0

, so

f(x) = (x-a) + (x-b) = 2x - (a+b)

, which increases with slope

2

. Therefore: on

(-\infty, a]

f

decreases (one-to-one); on

[a,b]

f

is constant (not one-to-one unless

a=b

); on

[b, \infty)

f

increases (one-to-one). Choice B is correct. Choice A works but choice B represents a cleaner interval. Choice C is wrong because constant functions aren't one-to-one. Choice D is incorrect because it spans both decreasing and constant regions.