IB Mathematics: Analysis and Approaches Quiz: Function Properties And Inverses

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Function Properties And InversesQuestion 1 of 20

A function $f(x)$ is defined as $f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}$ . Which property does this function have?

It is an even function.

It is neither even nor odd.

It is a periodic function.

It is an odd function.

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Function Properties And Inverses

Practice Function Properties And Inverses in IB Mathematics: Analysis and Approaches 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 Function Properties And Inverses, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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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.

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Question 1

A function $f(x)$ is defined as $f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}$ . Which property does this function have?

It is an even function.
It is neither even nor odd.
It is a periodic function.
It is an odd function. (correct answer)

Explanation: This question tests your understanding of even and odd functions, which is a key concept in function analysis. When you encounter piecewise functions like this, you need to check their symmetry properties by testing specific conditions. To determine if a function is odd, you must verify that

f(-x) = -f(x)

for all values in the domain. Let's test this systematically. For any positive value

x > 0

, we have

f(x) = x^2

and

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

. Also,

-f(x) = -x^2

. So

f(-x) = -f(x)

holds for positive inputs. For negative values, if

x < 0

, then

-x > 0

, so

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

and

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

. Again,

f(-x) = -f(x)

. Since this relationship holds for all values, the function is odd. Option A is incorrect because even functions satisfy

f(-x) = f(x)

, but here

f(1) = 1

while

f(-1) = -1

, so

f(-x) \neq f(x)

. Option B is wrong because we've proven the function is odd. Option C is incorrect because this function has no repeating pattern over regular intervals—it's not periodic. When checking function symmetry, always test both the even condition (

f(-x) = f(x)

) and odd condition (

f(-x) = -f(x)

) systematically. Remember that odd functions have rotational symmetry about the origin, while even functions have reflectional symmetry about the y-axis.

Question 2

A function $f(x)$ is defined as $f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}$ . Which property does this function have?

It is an even function.
It is neither even nor odd.
It is a periodic function.
It is an odd function. (correct answer)

f(-x) = -f(x)

for all values in the domain. Let's test this systematically. For any positive value

x > 0

, we have

f(x) = x^2

and

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

. Also,

-f(x) = -x^2

. So

f(-x) = -f(x)

holds for positive inputs. For negative values, if

x < 0

, then

-x > 0

, so

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

and

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

. Again,

f(-x) = -f(x)

. Since this relationship holds for all values, the function is odd. Option A is incorrect because even functions satisfy

f(-x) = f(x)

, but here

f(1) = 1

while

f(-1) = -1

, so

f(-x) \neq f(x)

f(-x) = f(x)

) and odd condition (

f(-x) = -f(x)

) systematically. Remember that odd functions have rotational symmetry about the origin, while even functions have reflectional symmetry about the y-axis.

Question 3

Given that $f(x)$ is an odd function and $\int_0^3 f(x) \,dx = 5$ , what is the value of $\int_{-3}^3 (2f(x) + 3) \,dx$ ?

10
18 (correct answer)
28
38

Explanation: We can split the integral using the property of linearity:

\int_{-3}^3 (2f(x) + 3) \,dx = \int_{-3}^3 2f(x) \,dx + \int_{-3}^3 3 \,dx = 2\int_{-3}^3 f(x) \,dx + \int_{-3}^3 3 \,dx

A key property of odd functions is that their definite integral over a symmetric interval

[-a, a]

is zero. Since

f(x)

is odd,

\int_{-3}^3 f(x) \,dx = 0

. The given information

\int_0^3 f(x) \,dx = 5

is consistent with this, as

\int_{-3}^0 f(x) \,dx = -5

. So, the first term is

2 \times 0 = 0

. The second term is the integral of a constant:

\int_{-3}^3 3 \,dx = [3x]_{-3}^3 = 3(3) - 3(-3) = 9 - (-9) = 18

The total value is

0 + 18 = 18

. Distractor A calculates

2 \times 5

and ignores the rest. Distractor C incorrectly computes

2\int_0^3 f(x)dx + 18 = 2(5)+18=28

. Distractor D incorrectly treats

f(x)

as an even function, where

\int_{-3}^3 f(x) dx = 2\int_0^3 f(x) dx = 10

, leading to

2(10) + 18 = 38

Question 4

Let $f(x)$ be a non-zero odd function and $g(x)$ be a non-zero even function, both defined for all $x \in \mathbb{R}$ . Let $h(x) = f(x) + g(x)$ . Which expression is always equivalent to $h(-x)$ ?

$h(x)$
$-h(x)$
$g(x) - f(x)$ (correct answer)
$f(x) - g(x)$

Explanation: By definition, an odd function satisfies

f(-x) = -f(x)

and an even function satisfies

g(-x) = g(x)

. We can substitute these into the expression for

h(-x)

h(-x) = f(-x) + g(-x)

h(-x) = -f(x) + g(x)

Rearranging the terms gives:

h(-x) = g(x) - f(x)

Therefore, option C is the correct answer. Distractor A is incorrect because it implies

h(x)

is even, which is not generally true. Distractor B is incorrect because it implies

h(x)

is odd, which is also not generally true. Distractor D is incorrect as it results from a sign error in the substitution.

Question 5

Let $f(x) = e^{2x}$ . The graph of $f$ is restricted to the domain $x \ge 0$ . Its inverse is $g(x)$ . What is $g'(e^2)$ ?

$\frac{1}{2e^2}$ (correct answer)
$\frac{1}{e^2}$
$\frac{e^2}{2}$
$2e^2$

Explanation: We can solve this in two ways. Method 1: Using the formula

g'(y_0) = \frac{1}{f'(x_0)}

where

g=f^{-1}

and

y_0 = f(x_0)

. Here

y_0 = e^2

. We need to find

x_0

such that

f(x_0) = e^2

e^{2x_0} = e^2 \implies 2x_0 = 2 \implies x_0 = 1

Now find the derivative of

f(x)

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

. Evaluate at

x_0=1

f'(1) = 2e^{2(1)} = 2e^2

. Therefore,

g'(e^2) = \frac{1}{f'(1)} = \frac{1}{2e^2}

. Method 2: Find the inverse function explicitly. Let

y = e^{2x}

. Swap variables:

x = e^{2y}

. Solve for

y

\ln x = 2y \implies y = \frac{1}{2}\ln x

. So

g(x) = \frac{1}{2}\ln x

. Now find the derivative of

g(x)

g'(x) = \frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2x}

. Evaluate at

x = e^2

g'(e^2) = \frac{1}{2e^2}

. Both methods yield the same result. Distractor D is

f'(1)

, forgetting the reciprocal. Distractor B misses the factor of 2 from the chain rule. Distractor C is an algebraic error.

Question 6

Let $f(x)$ be a non-zero odd function and $g(x)$ be a non-zero even function, both defined for all $x \in \mathbb{R}$ . Let $h(x) = f(x) + g(x)$ . Which expression is always equivalent to $h(-x)$ ?

$h(x)$
$-h(x)$
$g(x) - f(x)$ (correct answer)
$f(x) - g(x)$

Explanation: By definition, an odd function satisfies

f(-x) = -f(x)

and an even function satisfies

g(-x) = g(x)

. We can substitute these into the expression for

h(-x)

h(-x) = f(-x) + g(-x)

h(-x) = -f(x) + g(x)

Rearranging the terms gives:

h(-x) = g(x) - f(x)

Therefore, option C is the correct answer. Distractor A is incorrect because it implies

h(x)

is even, which is not generally true. Distractor B is incorrect because it implies

h(x)

is odd, which is also not generally true. Distractor D is incorrect as it results from a sign error in the substitution.

Question 7

Let $f(x) = \ln(x)$ for $x>0$ and $g(x) = x^2+1$ . The function $h(x) = (f \circ g)(x)$ is defined on a restricted domain $D$ to be one-to-one. Which of the following could be a valid expression for $h^{-1}(x)$ ?

$\sqrt{e^x - 1}$ (correct answer)
$e^{\sqrt{x-1}}$
$e^{x^2} - 1$
$\pm \sqrt{e^x - 1}$

Explanation: First, find the composite function

h(x)

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

. The natural domain is

x \in \mathbb{R}

since

x^2+1 > 0

. The function

h(x)

is even and therefore not one-to-one. To make it one-to-one, we must restrict its domain, for example, to

D=[0, \infty)

. On this domain, the range of

h(x)

[\ln(1), \infty) = [0, \infty)

. To find the inverse, let

y = \ln(x^2+1)

and swap variables:

x = \ln(y^2+1)

. Now, solve for

y

e^x = y^2+1

y^2 = e^x - 1

y = \pm\sqrt{e^x - 1}

The range of

h^{-1}(x)

must be the restricted domain of

h(x)

, which we chose as

D=[0, \infty)

. Therefore, we must choose the positive root:

h^{-1}(x) = \sqrt{e^x - 1}

. Distractor B confuses the order of operations. Distractor C has algebraic errors. Distractor D is not a function, and an inverse must be a function.

Question 8

$\sqrt{e^x - 1}$ (correct answer)
$e^{\sqrt{x-1}}$
$e^{x^2} - 1$
$\pm \sqrt{e^x - 1}$

Explanation: First, find the composite function

h(x)

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

. The natural domain is

x \in \mathbb{R}

since

x^2+1 > 0

. The function

h(x)

is even and therefore not one-to-one. To make it one-to-one, we must restrict its domain, for example, to

D=[0, \infty)

. On this domain, the range of

h(x)

[\ln(1), \infty) = [0, \infty)

. To find the inverse, let

y = \ln(x^2+1)

and swap variables:

x = \ln(y^2+1)

. Now, solve for

y

e^x = y^2+1

y^2 = e^x - 1

y = \pm\sqrt{e^x - 1}

The range of

h^{-1}(x)

must be the restricted domain of

h(x)

, which we chose as

D=[0, \infty)

. Therefore, we must choose the positive root:

h^{-1}(x) = \sqrt{e^x - 1}

. Distractor B confuses the order of operations. Distractor C has algebraic errors. Distractor D is not a function, and an inverse must be a function.

Question 9

Let $f(x) = x^3 + x - 2$ . The inverse function $f^{-1}(x)$ exists. Find the value of $(f^{-1})'(0)$ .

1
4
$\frac{1}{4}$ (correct answer)
$\frac{1}{3}$

Explanation: We use the formula for the derivative of an inverse function:

(f^{-1})'(y_0) = \frac{1}{f'(x_0)}

, where

y_0 = f(x_0)

. Here, we want to find

(f^{-1})'(0)

, so

y_0 = 0

. First, we need to find the value of

x_0

such that

f(x_0) = 0

x_0^3 + x_0 - 2 = 0

By inspection, we can see that

x_0 = 1

is the solution, since

1^3 + 1 - 2 = 0

. Next, we need to find the derivative of

f(x)

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

Now, we evaluate the derivative at

x_0 = 1

f'(1) = 3(1)^2 + 1 = 4

Finally, we apply the formula:

(f^{-1})'(0) = \frac{1}{f'(1)} = \frac{1}{4}

Distractor A is

f'(0)

. Distractor B is

f'(1)

, but forgets to take the reciprocal. Distractor D is an incorrect calculation of the derivative.

Question 10

Any function $f(x)$ can be written as the sum of an even function $f_e(x)$ and an odd function $f_o(x)$ . For the function $f(x) = e^x + \sin(x)$ , determine the even part, $f_e(x)$ .

$e^x$
$\cosh(x)$ (correct answer)
$\sinh(x)$
$\sinh(x) + \sin(x)$

Explanation: The even part of a function

f(x)

is given by the formula

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

. For

f(x) = e^x + \sin(x)

, we first find

f(-x)

f(-x) = e^{-x} + \sin(-x) = e^{-x} - \sin(x)

Now, we substitute into the formula for

f_e(x)

f_e(x) = \frac{(e^x + \sin(x)) + (e^{-x} - \sin(x))}{2}

f_e(x) = \frac{e^x + e^{-x}}{2}

This expression is the definition of the hyperbolic cosine function,

\cosh(x)

. Distractor A is part of the original function, not its even part. Distractor C is

\sinh(x)

, which is the even part of just

e^x

. Distractor D is the odd part of

f(x)

, which is

f_o(x) = \frac{f(x) - f(-x)}{2}

Question 11

The function $f(x) = \frac{ax+3}{2x-4}$ , where $x \neq 2$ , is self-inverse. Find the value of the constant $a$ .

-4
-2
2
4 (correct answer)

Explanation: A function

f

is self-inverse if

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

. First, we find the inverse of

f(x)

. Let

y = \frac{ax+3}{2x-4}

. To find the inverse, we swap

x

and

y

and solve for

y

x = \frac{ay+3}{2y-4}

x(2y-4) = ay+3

2xy - 4x = ay+3

2xy - ay = 4x+3

y(2x - a) = 4x+3

y = \frac{4x+3}{2x-a}

So,

f^{-1}(x) = \frac{4x+3}{2x-a}

. For

f

to be self-inverse, we must have

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

\frac{ax+3}{2x-4} = \frac{4x+3}{2x-a}

By comparing coefficients, we see that we must have

a=4

. This also makes the denominators consistent, as

-4 = -a

also gives

a=4

. Distractor A results from the common error of setting

a

equal to the constant in the denominator,

-4

. Distractors B and C confuse coefficients.

Question 12

Let $f$ be a non-zero even function and $g$ be a non-zero odd function. Which of the following statements must be true?

$(f \circ g)$ is an odd function.
$(g \circ f)$ is an even function. (correct answer)
$f(0) = g(0)$
$f+g$ is an odd function.

Explanation: Let's test each statement using the definitions

f(-x) = f(x)

(even) and

g(-x) = -g(x)

(odd). A:

(f \circ g)(-x) = f(g(-x)) = f(-g(x))

. Since

f

is even,

f(-y) = f(y)

. So,

f(-g(x)) = f(g(x)) = (f \circ g)(x)

. This shows

f \circ g

is an even function, so A is false. B:

(g \circ f)(-x) = g(f(-x))

. Since

f

is even,

f(-x) = f(x)

. So,

g(f(-x)) = g(f(x)) = (g \circ f)(x)

. This shows

g \circ f

is an even function, so B is true. C: For an odd function

g

g(0) = g(-0) = -g(0)

, which implies

2g(0)=0

and so

g(0)=0

. For an even function

f

, there is no restriction on the value of

f(0)

. For example,

f(x) = x^2+1

is even but

f(0)=1

. So

f(0)=g(0)

is not always true. C is false. D:

(f+g)(-x) = f(-x)+g(-x) = f(x) - g(x)

. This is not equal to

(f+g)(x)

-(f+g)(x)

in general, so

f+g

is neither odd nor even (unless one function is zero, which is ruled out). D is false.

Question 13

The function $f(x) = (x-c)^2 + d$ is restricted to the domain $x \ge c$ . Its inverse is $f^{-1}(x) = 3 + \sqrt{x-5}$ . What are the values of $c$ and $d$ ?

$c=3, d=5$ (correct answer)
$c=5, d=3$
$c=-3, d=-5$
$c=3, d=-5$

Explanation: Let's find the inverse of the general function

f(x)=(x-c)^2+d

with domain

x \ge c

. The range is

y \ge d

. Let

y=(x-c)^2+d

. Swap variables:

x=(y-c)^2+d

. Solve for

y

x-d=(y-c)^2 \implies y-c = \pm\sqrt{x-d} \implies y = c \pm\sqrt{x-d}

. The range of the inverse must be the domain of the original, which is

y \ge c

. So we must choose the positive root:

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

. We are given that

f^{-1}(x) = 3 + \sqrt{x-5}

. By comparing the two forms, we can equate the corresponding parts:

c = 3

-d = -5 \implies d = 5

. Therefore,

c=3

and

d=5

. Distractor B confuses the values of c and d. Distractors C and D involve sign errors.

Question 14

The function $f(x) = 3\cos(2x) - 1$ is restricted to the domain $D = [0, \frac{\pi}{2}]$ to ensure it is one-to-one. What is the domain of the inverse function $f^{-1}(x)$ ?

$[-4, 2]$ (correct answer)
$[-2, 4]$
$[-1, 1]$
$[0, \frac{\pi}{2}]$

Explanation: The domain of the inverse function

f^{-1}(x)

is the range of the original function

f(x)

on its restricted domain. The restricted domain of

f(x)

x \in [0, \frac{\pi}{2}]

. For this interval of

x

, the argument of the cosine function,

2x

, is in the interval

[0, \pi]

. Over the interval

[0, \pi]

, the function

\cos(u)

takes all values from

\cos(0)=1

down to

\cos(\pi)=-1

. So the range of

\cos(2x)

on the given domain is

[-1, 1]

. Now we find the range of

f(x) = 3\cos(2x) - 1

. The maximum value is

3(1) - 1 = 2

. The minimum value is

3(-1) - 1 = -4

. So, the range of

f(x)

on the domain

D

[-4, 2]

. This is the domain of

f^{-1}(x)

. Distractor B comes from a sign error

1 \pm 3

. Distractor C is the range of

\cos(2x)

, not

f(x)

. Distractor D is the range of

f^{-1}(x)

, not its domain.

Question 15

Given that $f(x) = 2x + \cos(x)$ is a one-to-one function, find the value of $(f^{-1})'(1)$ .

$\frac{1}{3}$
$\frac{1}{2}$ (correct answer)
1
2

Explanation: We use the formula for the derivative of an inverse function:

(f^{-1})'(y_0) = \frac{1}{f'(x_0)}

, where

y_0 = f(x_0)

. In this problem, we want to find

(f^{-1})'(1)

, so

y_0 = 1

. First, we need to find the value of

x_0

such that

f(x_0) = 1

2x_0 + \cos(x_0) = 1

By inspection, we can see that

x_0 = 0

is the solution, since

2(0) + \cos(0) = 0 + 1 = 1

. Next, we find the derivative of

f(x)

f'(x) = 2 - \sin(x)

Now, we evaluate the derivative at

x_0 = 0

f'(0) = 2 - \sin(0) = 2 - 0 = 2

Finally, we apply the formula:

(f^{-1})'(1) = \frac{1}{f'(0)} = \frac{1}{2}

Distractor D is the value of

f'(0)

, but forgets to take the reciprocal. Distractor C could be the value of

f(0)

. Distractor A is a plausible calculation error.

Question 16

The function $f(x)$ is periodic with a fundamental period of 3. A new function is defined as $g(x) = 2f(\frac{x}{4} - 1) + 5$ . What is the fundamental period of $g(x)$ ?

$\frac{3}{4}$
3
12 (correct answer)
13

Explanation: The period of a function is affected only by horizontal transformations (stretches, compressions, and reflections inside the function argument). Vertical shifts and stretches do not change the period. The transformation is of the form

Af(B(x-C))+D

. The new period

P'

is related to the old period

P

P' = P/|B|

. In this case,

f(x)

has period

P=3

. The argument of

f

g(x)

\frac{x}{4} - 1 = \frac{1}{4}(x-4)

. So,

B = 1/4

. The new period is

P' = \frac{3}{|1/4|} = 3 \times 4 = 12

. Distractor A results from incorrectly multiplying the period by

B

. Distractor B ignores the horizontal stretch. Distractor D incorrectly involves the horizontal shift in the period calculation.

Question 17

Let $f(x) = e^{2x}$ . The graph of $f$ is restricted to the domain $x \ge 0$ . Its inverse is $g(x)$ . What is $g'(e^2)$ ?

$\frac{1}{2e^2}$ (correct answer)
$\frac{1}{e^2}$
$\frac{e^2}{2}$
$2e^2$

Explanation: We can solve this in two ways. Method 1: Using the formula

g'(y_0) = \frac{1}{f'(x_0)}

where

g=f^{-1}

and

y_0 = f(x_0)

. Here

y_0 = e^2

. We need to find

x_0

such that

f(x_0) = e^2

e^{2x_0} = e^2 \implies 2x_0 = 2 \implies x_0 = 1

Now find the derivative of

f(x)

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

. Evaluate at

x_0=1

f'(1) = 2e^{2(1)} = 2e^2

. Therefore,

g'(e^2) = \frac{1}{f'(1)} = \frac{1}{2e^2}

. Method 2: Find the inverse function explicitly. Let

y = e^{2x}

. Swap variables:

x = e^{2y}

. Solve for

y

\ln x = 2y \implies y = \frac{1}{2}\ln x

. So

g(x) = \frac{1}{2}\ln x

. Now find the derivative of

g(x)

g'(x) = \frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2x}

. Evaluate at

x = e^2

g'(e^2) = \frac{1}{2e^2}

. Both methods yield the same result. Distractor D is

f'(1)

, forgetting the reciprocal. Distractor B misses the factor of 2 from the chain rule. Distractor C is an algebraic error.

Question 18

Any function $f(x)$ can be written as the sum of an even function $f_e(x)$ and an odd function $f_o(x)$ . For the function $f(x) = e^x + \sin(x)$ , determine the even part, $f_e(x)$ .

$e^x$
$\cosh(x)$ (correct answer)
$\sinh(x)$
$\sinh(x) + \sin(x)$

Explanation: The even part of a function

f(x)

is given by the formula

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

. For

f(x) = e^x + \sin(x)

, we first find

f(-x)

f(-x) = e^{-x} + \sin(-x) = e^{-x} - \sin(x)

Now, we substitute into the formula for

f_e(x)

f_e(x) = \frac{(e^x + \sin(x)) + (e^{-x} - \sin(x))}{2}

f_e(x) = \frac{e^x + e^{-x}}{2}

This expression is the definition of the hyperbolic cosine function,

\cosh(x)

. Distractor A is part of the original function, not its even part. Distractor C is

\sinh(x)

, which is the even part of just

e^x

. Distractor D is the odd part of

f(x)

, which is

f_o(x) = \frac{f(x) - f(-x)}{2}

Question 19

Let $f(x)$ be a differentiable odd function. Which of the following statements about its derivative, $f'(x)$ , must be true?

$f'(x)$ is an odd function.
$f'(x)$ is an even function. (correct answer)
$f'(0) = 0$ .
$f'(x)$ can be neither odd nor even.

Explanation: The definition of an odd function is

f(-x) = -f(x)

. To find the property of its derivative, we differentiate this identity with respect to

x

using the chain rule on the left side:

\frac{d}{dx}f(-x) = \frac{d}{dx}(-f(x))

f'(-x) \cdot \frac{d}{dx}(-x) = -f'(x)

f'(-x) \cdot (-1) = -f'(x)

-f'(-x) = -f'(x)

f'(-x) = f'(x)

This is the definition of an even function. Thus, the derivative of an odd function is an even function. Distractor A is incorrect; it applies to the derivative of an even function. Distractor C is not always true; consider

f(x) = \sin(x)

, which is odd. Its derivative is

f'(x) = \cos(x)

, and

f'(0) = \cos(0) = 1 \neq 0

. Distractor D is incorrect as we have proven

f'(x)

must be even.

Question 20

The function $f(x)$ is periodic with a fundamental period of 3. A new function is defined as $g(x) = 2f(\frac{x}{4} - 1) + 5$ . What is the fundamental period of $g(x)$ ?

$\frac{3}{4}$
3
12 (correct answer)
13

Af(B(x-C))+D

. The new period

P'

is related to the old period

P

P' = P/|B|

. In this case,

f(x)

has period

P=3

. The argument of

f

g(x)

\frac{x}{4} - 1 = \frac{1}{4}(x-4)

. So,

B = 1/4

. The new period is

P' = \frac{3}{|1/4|} = 3 \times 4 = 12

. Distractor A results from incorrectly multiplying the period by

B

. Distractor B ignores the horizontal stretch. Distractor D incorrectly involves the horizontal shift in the period calculation.