IB Mathematics: Analysis and Approaches Quiz: Function Composition

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Function CompositionQuestion 1 of 20

The radius $r$ of a circular ripple in a pond, in meters, is given by the function $r(t) = 10\sqrt{t}$ , where $t$ is the time in seconds after a stone is dropped. The area $A$ of a circle is given by $A(r) = \pi r^2$ . Which function $(A \circ r)(t)$ correctly represents the area of the ripple as a function of time?

10\pi\sqrt{t}

100\pi t

100\pi \sqrt{t}

100\pi t^2

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Function Composition

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

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

$10\pi\sqrt{t}$
$100\pi t$ (correct answer)
$100\pi \sqrt{t}$
$100\pi t^2$

Explanation: We need to find the composite function

(A \circ r)(t) = A(r(t))

.
This means we substitute the expression for

r(t)

into the function

A(r)

r(t) = 10\sqrt{t}\ \(A(r) = \pi r^2\ \((A \circ r)(t) = A(10\sqrt{t}) = \pi (10\sqrt{t})^2 = \pi (10^2 (\sqrt{t})^2) = \pi(100t) = 100\pi t

.
Distractor A incorrectly computes

\pi \cdot r(t)

. Distractor C comes from not squaring the

\sqrt{t}

term. Distractor D comes from squaring

t

instead of

\sqrt{t}

Question 2

$10\pi\sqrt{t}$
$100\pi t$ (correct answer)
$100\pi \sqrt{t}$
$100\pi t^2$

Explanation: We need to find the composite function

(A \circ r)(t) = A(r(t))

.
This means we substitute the expression for

r(t)

into the function

A(r)

r(t) = 10\sqrt{t}\ \(A(r) = \pi r^2\ \((A \circ r)(t) = A(10\sqrt{t}) = \pi (10\sqrt{t})^2 = \pi (10^2 (\sqrt{t})^2) = \pi(100t) = 100\pi t

.
Distractor A incorrectly computes

\pi \cdot r(t)

. Distractor C comes from not squaring the

\sqrt{t}

term. Distractor D comes from squaring

t

instead of

\sqrt{t}

Question 3

Let $f(x)=x^2$ , $g(x)=x+2$ , and $h(x)=2x$ . Find the sum of the values of $x$ for which $(f \circ g)(x) = (h \circ f)(x)$ .

$-4$
$0$
$2$
$4$ (correct answer)

Explanation: First, find the expressions for the two composite functions.

(f \circ g)(x) = f(g(x)) = f(x+2) = (x+2)^2 = x^2 + 4x + 4

(h \circ f)(x) = h(f(x)) = h(x^2) = 2(x^2) = 2x^2

.
Now, set the two expressions equal to each other:

x^2 + 4x + 4 = 2x^2\ Rearrange to form a standard quadratic equation:\ \(0 = 2x^2 - x^2 - 4x - 4\ \(0 = x^2 - 4x - 4\ Let the roots of this quadratic equation be \(\alpha

and

\beta

. The sum of the roots of a quadratic equation

ax^2+bx+c=0

is given by

-b/a

.
In this case,

a=1, b=-4, c=-4

.
The sum of the values of

x

-(-4)/1 = 4

.
Alternatively, one could solve for

x

using the quadratic formula, get

x = 2 \pm 2\sqrt{2}

, and then sum the roots:

(2+2\sqrt{2}) + (2-2\sqrt{2}) = 4

.
Distractor A results from a sign error in the sum of roots formula (using

b/a

). Distractor B would be the sum if the equation was

x^2-4=0

Question 4

Given that $(f \circ g)(x) = 4x^2 - 12x + 9$ and $g(x) = 2x - 3$ , which of the following is a possible expression for $f(x)$ ?

$f(x) = x^2$ (correct answer)
$f(x) = 2x+1$
$f(x) = 2x^2+3x$
$f(x) = 4x^2 - 12x + 9$

Explanation: We have

(f \circ g)(x) = f(g(x)) = f(2x-3)

.
We are given

f(2x-3) = 4x^2 - 12x + 9

.
We can try to express the right-hand side in terms of

g(x) = 2x-3

.
Notice that

4x^2 - 12x + 9

is a perfect square trinomial:

(2x-3)^2

.
So, we have

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

.
Let

u = 2x-3

. Then the equation becomes

f(u) = u^2

.
This means the function

f

squares its input. Therefore,

f(x) = x^2

.
Distractor D implies that

f

is the identity function, which is incorrect. Distractors B and C, if composed with

g(x)

, would not produce the given expression.

Question 5

Let $f(x) = \frac{1}{x-1}$ and $g(x) = \frac{x}{x+1}$ . Find a simplified expression for $(f \circ g)(x)$ .

$\frac{1}{x}$
$x$
$x+1$
$-x-1$ (correct answer)

Explanation: To find

(f \circ g)(x)

, we substitute

g(x)

into

f(x)

(f \circ g)(x) = f(g(x)) = f\left(\frac{x}{x+1}\right) = \frac{1}{\frac{x}{x+1} - 1}\ To simplify the expression, we work with the denominator:\ \(\frac{x}{x+1} - 1 = \frac{x}{x+1} - \frac{x+1}{x+1} = \frac{x - (x+1)}{x+1} = \frac{x - x - 1}{x+1} = \frac{-1}{x+1}\ Now substitute this back into the main expression:\ \((f \circ g)(x) = \frac{1}{\frac{-1}{x+1}} = 1 \cdot \frac{x+1}{-1} = -(x+1) = -x-1

.
Distractor A (1/x) is the result of composing in the wrong order:

(g \circ f)(x) = g(\frac{1}{x-1}) = \frac{1/(x-1)}{1/(x-1) + 1} = \frac{1}{1 + (x-1)} = \frac{1}{x}

. Distractors B and C are from algebraic errors during simplification.

Question 6

Let $f(x) = 2 - \frac{1}{x}$ . Find the value of $(f \circ f \circ f)(2)$ .

$3/2$
$4/3$
$5/4$ (correct answer)
$2$

Explanation: We need to compute the function composition iteratively.
First, find

f(2)

f(2) = 2 - \frac{1}{2} = \frac{3}{2}\ Next, find \((f \circ f)(2) = f(f(2)) = f(\frac{3}{2})\ \(f(\frac{3}{2}) = 2 - \frac{1}{3/2} = 2 - \frac{2}{3} = \frac{6-2}{3} = \frac{4}{3}\ Finally, find \((f \circ f \circ f)(2) = f(f(f(2))) = f(\frac{4}{3})\ \(f(\frac{4}{3}) = 2 - \frac{1}{4/3} = 2 - \frac{3}{4} = \frac{8-3}{4} = \frac{5}{4}\ Distractors A and B are the results of the first and second iterations, respectively. Distractor D would only be correct if \(f(x)=x

or if the function value cycled back to the start.

Question 7

Let $f(x) = 2x + 3$ and $g(x) = \begin{cases} x^2 & \text{if } x \ge 1 \\ 2x-1 & \text{if } x < 1 \end{cases}$ . Find the value of $(g \circ f)(-2)$ .

$-7$
$-3$ (correct answer)
$1$
$9$

Explanation: To find

(g \circ f)(-2)

, we first compute the inner function

f(-2)

f(-2) = 2(-2) + 3 = -4 + 3 = -1

.
Now we compute

g(f(-2))

, which is

g(-1)

.
To evaluate

g(-1)

, we must use the rule for the piecewise function

g(x)

. Since

-1 < 1

, we use the second piece:

g(x) = 2x-1

g(-1) = 2(-1) - 1 = -2 - 1 = -3

.
Therefore,

(g \circ f)(-2) = -3

.
Distractor C (1) arises from using the wrong piece of the definition for

g(x)

, calculating

g(-1) = (-1)^2 = 1

. Distractor A (-7) comes from composing in the wrong order:

g(-2) = 2(-2)-1=-5

, and then

f(-5)=2(-5)+3 = -7

. Distractor D (9) would come from

(f \circ g)(-2)

, where

g(-2)=-5

, but with an error

f(-5)=2(-5)+3 = -7

. Or maybe

g(-2)=-3

error, then

f(-3)=-3

. No, wait. Maybe

g(-2)=(-2)^2=4

, then

f(4)=11

. A distractor of 1 is very plausible.

Question 8

Let $f(x) = x^2+4$ and $g(x) = \sin(x)$ . Find the range of the composite function $(f \circ g)(x)$ .

$[4, 5]$ (correct answer)
$[0, 5]$
$[4, \infty)$
$[5, \infty)$

Explanation: The composite function is

(f \circ g)(x) = f(g(x)) = f(\sin(x)) = (\sin(x))^2 + 4

.
To find the range of the composite function, we first consider the range of the inner function,

g(x) = \sin(x)

. The range of

\sin(x)

[-1, 1]

.
This range becomes the effective domain for the outer function

f(x)

. We need to find the range of

f(u) = u^2+4

for

u \in [-1, 1]

.
The function

f(u) = u^2+4

is a parabola opening upwards with its vertex at

u=0

.
The minimum value on the interval

[-1, 1]

occurs at the vertex,

u=0

, where

f(0) = 0^2+4 = 4

.
The maximum value occurs at the endpoints of the interval,

u=-1

and

u=1

f(1) = 1^2+4 = 5

and

f(-1) = (-1)^2+4 = 5

.
Therefore, the range of

(f \circ g)(x)

[4, 5]

.
Distractor C is the range of

f(x)

for all real

x

. Distractor D results from only checking the endpoints

u=\pm 1

and ignoring the minimum at the vertex. Distractor B may result from confusion about the range of

\sin^2(x)

and combining it with the range of

f(x)

Question 9

Let $f(x) = x^3$ and $g(x) = x + k$ . Given that $(f \circ g)(2) = 27$ , find the value of $k$ .

$-5$
$1$ (correct answer)
$5$
$19$

Explanation: First, evaluate the composite function at

x=2

(f \circ g)(2) = f(g(2))

.
Find the value of the inner function

g(2)

g(2) = 2 + k

.
Now substitute this result into

f(x)

f(g(2)) = f(2+k) = (2+k)^3

.
We are given that this value is 27:

(2+k)^3 = 27\ Take the cube root of both sides:\ \(2+k = 3\ \(k = 3 - 2 = 1

.
Distractor D (19) comes from composing in the wrong order:

(g \circ f)(2) = g(f(2)) = g(2^3) = g(8) = 8+k

. Setting

8+k = 27

gives

k=19

. Distractor A (-5) arises from incorrectly taking the cube root:

2+k=-3

. Distractor C is a simple arithmetic error.

Question 10

Let $f(x) = x^2 + 1$ and $g(x) = 2x - 1$ . Find the values of $x$ for which $(f \circ g)(x) = (g \circ f)(x)$ .

$x = 1 \pm \frac{\sqrt{2}}{2}$ (correct answer)
$x = 1 \pm \sqrt{2}$
$x = 2 \pm \frac{\sqrt{2}}{2}$
$x = 2 \pm \sqrt{2}$

Explanation: First, find the expressions for the composite functions.

(f \circ g)(x) = f(g(x)) = f(2x - 1) = (2x - 1)^2 + 1 = (4x^2 - 4x + 1) + 1 = 4x^2 - 4x + 2\ \((g \circ f)(x) = g(f(x)) = g(x^2 + 1) = 2(x^2 + 1) - 1 = 2x^2 + 2 - 1 = 2x^2 + 1\ Now, set them equal to each other:\ \(4x^2 - 4x + 2 = 2x^2 + 1\ \(2x^2 - 4x + 1 = 0\ This is a quadratic equation. Since it does not factor easily, use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)}\ \(x = \frac{4 \pm \sqrt{16 - 8}}{4}\ \(x = \frac{4 \pm \sqrt{8}}{4}\ \(x = \frac{4 \pm 2\sqrt{2}}{4}\ \(x = 1 \pm \frac{2\sqrt{2}}{4}\ \(x = 1 \pm \frac{\sqrt{2}}{2}

Distractor B comes from an error in simplifying

\frac{4 \pm 2\sqrt{2}}{4}

1 \pm \sqrt{2}

. Distractors C and D arise from calculation errors in the quadratic formula.

Question 11

Let $f(x) = 2x - 5$ and $g(x) = ax + b$ . The composite function $(f \circ g)(x)$ is a constant function with a value of 7 for all $x$ . Find the value of $b$ .

$0$
$2.5$
$3.5$
$6$ (correct answer)

Explanation: First, find the expression for the composite function

(f \circ g)(x)

(f \circ g)(x) = f(g(x)) = f(ax+b) = 2(ax+b) - 5 = 2ax + 2b - 5

.
We are told this is a constant function. For a linear function

mx+c

to be constant, the slope

m

must be zero. In our expression

2ax + (2b-5)

, the coefficient of

x

2a

.
So, we must have

2a = 0

, which implies

a=0

.
This means the function simplifies to

(f \circ g)(x) = 2(0)x + 2b - 5 = 2b - 5

.
We are given that the constant value is 7. Therefore:

2b - 5 = 7\ \(2b = 12\ \(b = 6

.
Distractor A is the value of

a

. Distractor B is the value of

b

2b-5=0

. Distractor C is the value of

b

2b=7

Question 12

Let $f(x)$ be an even function and $g(x)$ be an odd function. Which of the following composite functions is always an even function?

$(f \circ g)(x)$ only
$(g \circ f)(x)$ only
Both $(f \circ g)(x)$ and $(g \circ f)(x)$ (correct answer)
Neither $(f \circ g)(x)$ nor $(g \circ f)(x)$

Explanation: Recall the definitions:

f

is even if

f(-x) = f(x)

for all

x

g

is odd if

g(-x) = -g(x)

for all

x

.
Let's test

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

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

Since

g

is odd,

g(-x) = -g(x)

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

Since

f

is even,

f(-u) = f(u)

. Let

u=g(x)

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

.
Therefore,

(f \circ g)(x)

is an even function.
Now let's test

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

k(-x) = g(f(-x))

Since

f

is even,

f(-x) = f(x)

k(-x) = g(f(x)) = k(x)

.
Therefore,

(g \circ f)(x)

is also an even function.
Since both are always even, the correct option is C.

Question 13

Let $f(x) = \cos(x)$ and $g(x) = 2x + \frac{\pi}{2}$ . Find the exact value of $(f \circ g)(\frac{\pi}{3})$ .

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

Explanation: We need to calculate

(f \circ g)(\frac{\pi}{3}) = f(g(\frac{\pi}{3}))

.
First, evaluate the inner function

g(\frac{\pi}{3})

g(\frac{\pi}{3}) = 2(\frac{\pi}{3}) + \frac{\pi}{2} = \frac{2\pi}{3} + \frac{\pi}{2} = \frac{4\pi}{6} + \frac{3\pi}{6} = \frac{7\pi}{6}

.
Now, evaluate the outer function

f

with this result:

f(\frac{7\pi}{6}) = \cos(\frac{7\pi}{6})

.
The angle

\frac{7\pi}{6}

is in the third quadrant. The reference angle is

\frac{7\pi}{6} - \pi = \frac{\pi}{6}

.
We know

\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}

.
Since cosine is negative in the third quadrant,

\cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2}

.
Distractor D ignores the negative sign for cosine in the third quadrant. Distractor B is the value of

\sin(\frac{7\pi}{6})

. Distractor C has both a sign error and uses the wrong trigonometric function.

Question 14

The function $f(x) = \frac{ax+3}{x-2}$ , where $a$ is a constant, is a self-inverse function. This means that $(f \circ f)(x) = x$ for all $x$ in the domain. Find the value of $a$ .

$-3$
$-2$
$2$ (correct answer)
$3$

Explanation: For a function to be self-inverse, we need

f(f(x)) = x

. Let's compute

f(f(x))

f(f(x)) = f\left(\frac{ax+3}{x-2}\right) = \frac{a\left(\frac{ax+3}{x-2}\right)+3}{\left(\frac{ax+3}{x-2}\right)-2}

\Simplifying the numerator:

a\left(\frac{ax+3}{x-2}\right)+3 = \frac{a(ax+3)+3(x-2)}{x-2} = \frac{a^2x+3a+3x-6}{x-2}

\Simplifying the denominator:

\left(\frac{ax+3}{x-2}\right)-2 = \frac{ax+3-2(x-2)}{x-2} = \frac{ax+3-2x+4}{x-2} = \frac{(a-2)x+7}{x-2}

\So

f(f(x)) = \frac{a^2x+3a+3x-6}{(a-2)x+7}

\For this to equal

x

, we need

\frac{a^2x+3a+3x-6}{(a-2)x+7} = x

\Cross-multiplying:

a^2x+3a+3x-6 = x((a-2)x+7) = (a-2)x^2+7x

\Rearranging:

(a-2)x^2+7x-a^2x-3a-3x+6 = 0

(a-2)x^2+(4-a^2)x+(6-3a) = 0

\For this to hold for all

x

, all coefficients must be zero:\

a-2 = 0 \Rightarrow a = 2

\We can verify: when

a=2

, the coefficient of

x^2

0

, the coefficient of

x

4-4=0

, and the constant term is

6-6=0

Question 15

Let $f(x) = x^2 + 1$ and $g(x) = 2x - 1$ . Find the values of $x$ for which $(f \circ g)(x) = (g \circ f)(x)$ .

$x = 1 \pm \frac{\sqrt{2}}{2}$ (correct answer)
$x = 1 \pm \sqrt{2}$
$x = 2 \pm \frac{\sqrt{2}}{2}$
$x = 2 \pm \sqrt{2}$

Explanation: First, find the expressions for the composite functions.

(f \circ g)(x) = f(g(x)) = f(2x - 1) = (2x - 1)^2 + 1 = (4x^2 - 4x + 1) + 1 = 4x^2 - 4x + 2\ \((g \circ f)(x) = g(f(x)) = g(x^2 + 1) = 2(x^2 + 1) - 1 = 2x^2 + 2 - 1 = 2x^2 + 1\ Now, set them equal to each other:\ \(4x^2 - 4x + 2 = 2x^2 + 1\ \(2x^2 - 4x + 1 = 0\ This is a quadratic equation. Since it does not factor easily, use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)}\ \(x = \frac{4 \pm \sqrt{16 - 8}}{4}\ \(x = \frac{4 \pm \sqrt{8}}{4}\ \(x = \frac{4 \pm 2\sqrt{2}}{4}\ \(x = 1 \pm \frac{2\sqrt{2}}{4}\ \(x = 1 \pm \frac{\sqrt{2}}{2}

Distractor B comes from an error in simplifying

\frac{4 \pm 2\sqrt{2}}{4}

1 \pm \sqrt{2}

. Distractors C and D arise from calculation errors in the quadratic formula.

Question 16

The function $h(x) = 3(x+2)^2 - 5$ can be written as a composition $(f \circ g)(x)$ . If $g(x) = x+2$ , what is $f(x)$ ?

$f(x) = 3x^2 - 1$
$f(x) = 3x^2 - 5$ (correct answer)
$f(x) = 3(x-2)^2 - 5$
$f(x) = (3x-5)^2+2$

Explanation: We are given

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

.
We have

h(x) = 3(x+2)^2 - 5

and

g(x) = x+2

.
Substituting

g(x)

into the expression for

h(x)

gives

h(x) = 3(g(x))^2 - 5

.
Since

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

, we can see that

f(g(x)) = 3(g(x))^2 - 5

.
This means the rule for the function

f

is to take its input, square it, multiply by 3, and then subtract 5. Therefore,

f(x) = 3x^2 - 5

.
Alternatively, let

u = g(x) = x+2

. Then

x = u-2

. Substitute this into the expression for

h(x)

f(u) = 3((u-2)+2)^2 - 5 = 3u^2 - 5

. So

f(x) = 3x^2 - 5

.
Distractor C results from incorrectly substituting

x = u+2

instead of

x=u-2

. Distractor A and D represent other algebraic mistakes.

Question 17

Let $f(x) = e^{x-1}$ and $g(x) = \ln(x) + 1$ . Find the value of $(g \circ f)(5) + (f \circ g)(e)$ .

$5$
$e$
$6$
$5+e$ (correct answer)

Explanation: A student might notice that

f(x)

and

g(x)

are inverse functions. To verify, find the inverse of

f(x)

: let

y = e^{x-1}

. Then

\ln(y) = x-1

, so

x = \ln(y) + 1

. Thus,

f^{-1}(x) = \ln(x) + 1 = g(x)

.
By the property of inverse functions,

(g \circ f)(x) = (f^{-1} \circ f)(x) = x

and

(f \circ g)(x) = (f \circ f^{-1})(x) = x

.
Therefore,

(g \circ f)(5) = 5

and

(f \circ g)(e) = e

.
The sum is

5 + e

.
Alternatively, one can compute each part directly:

f(5) = e^{5-1} = e^4

(g \circ f)(5) = g(e^4) = \ln(e^4) + 1 = 4+1 = 5

g(e) = \ln(e) + 1 = 1+1 = 2

(f \circ g)(e) = f(2) = e^{2-1} = e^1 = e

.
The sum is

5+e

Question 18

Let $f(x) = |x - 2|$ and $g(x) = 3x + 1$ . Find the set of values of $x$ for which $(f \circ g)(x) < 5$ .

$-\frac{4}{3} < x < 2$ (correct answer)
$\frac{2}{3} < x < \frac{10}{3}$
$x < -\frac{4}{3} \text{ or } x > 2$
$x < 2$

Explanation: First, find the composite function

(f \circ g)(x)

(f \circ g)(x) = f(g(x)) = f(3x+1) = |(3x+1) - 2| = |3x - 1|

.
Now, we solve the inequality

|3x - 1| < 5

.
This absolute value inequality is equivalent to the double inequality:

-5 < 3x - 1 < 5\ Add 1 to all parts:\ \(-4 < 3x < 6\ Divide by 3:\ \(-\frac{4}{3} < x < 2\ Distractor B comes from composing in the wrong order: \((g \circ f)(x) = g(|x-2|) = 3|x-2|+1

. Solving

3|x-2|+1 < 5

gives

|x-2| < 4/3

, which leads to

2/3 < x < 10/3

. Distractor C comes from misinterpreting the absolute value inequality as

>

instead of

<

. Distractor D comes from only solving one side of the inequality,

3x-1 < 5

Question 19

Let $f(x) = 2x - 5$ and $g(x) = ax + b$ . The composite function $(f \circ g)(x)$ is a constant function with a value of 7 for all $x$ . Find the value of $b$ .

$0$
$2.5$
$3.5$
$6$ (correct answer)

Explanation: First, find the expression for the composite function

(f \circ g)(x)

(f \circ g)(x) = f(g(x)) = f(ax+b) = 2(ax+b) - 5 = 2ax + 2b - 5

.
We are told this is a constant function. For a linear function

mx+c

to be constant, the slope

m

must be zero. In our expression

2ax + (2b-5)

, the coefficient of

x

2a

.
So, we must have

2a = 0

, which implies

a=0

.
This means the function simplifies to

(f \circ g)(x) = 2(0)x + 2b - 5 = 2b - 5

.
We are given that the constant value is 7. Therefore:

2b - 5 = 7\ \(2b = 12\ \(b = 6

.
Distractor A is the value of

a

. Distractor B is the value of

b

2b-5=0

. Distractor C is the value of

b

2b=7

Question 20

Let $f(x)=x^2$ , $g(x)=x+2$ , and $h(x)=2x$ . Find the sum of the values of $x$ for which $(f \circ g)(x) = (h \circ f)(x)$ .

$-4$
$0$
$2$
$4$ (correct answer)

Explanation: First, find the expressions for the two composite functions.

(f \circ g)(x) = f(g(x)) = f(x+2) = (x+2)^2 = x^2 + 4x + 4

(h \circ f)(x) = h(f(x)) = h(x^2) = 2(x^2) = 2x^2

.
Now, set the two expressions equal to each other:

x^2 + 4x + 4 = 2x^2\ Rearrange to form a standard quadratic equation:\ \(0 = 2x^2 - x^2 - 4x - 4\ \(0 = x^2 - 4x - 4\ Let the roots of this quadratic equation be \(\alpha

and

\beta

. The sum of the roots of a quadratic equation

ax^2+bx+c=0

is given by

-b/a

.
In this case,

a=1, b=-4, c=-4

.
The sum of the values of

x

-(-4)/1 = 4

.
Alternatively, one could solve for

x

using the quadratic formula, get

x = 2 \pm 2\sqrt{2}

, and then sum the roots:

(2+2\sqrt{2}) + (2-2\sqrt{2}) = 4

.
Distractor A results from a sign error in the sum of roots formula (using

b/a

). Distractor B would be the sum if the equation was

x^2-4=0