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

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Function Notation And InversesQuestion 1 of 20

Given $f(x) = 2x+1$ and $g(x) = x^3$ , find the value of $(g \circ f^{-1})(9)$ .

364

729

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IB Mathematics: Analysis and Approaches Quiz

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

Practice Function Notation 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.

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This quiz focuses on Function Notation 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

Given $f(x) = 2x+1$ and $g(x) = x^3$ , find the value of $(g \circ f^{-1})(9)$ .

4
64 (correct answer)
364
729

Explanation: The expression

(g \circ f^{-1})(9)

means

g(f^{-1}(9))

. First, we need to find the value of

f^{-1}(9)

. Let

f^{-1}(9) = k

. By the definition of an inverse function,

f(k) = 9

. So,

2k+1 = 9

, which gives

2k=8

and

k=4

. Thus,

f^{-1}(9) = 4

. Now, we substitute this result into

g

g(f^{-1}(9)) = g(4)

. Since

g(x) = x^3

g(4) = 4^3 = 64

Question 2

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

-3
-2
2 (correct answer)
3

Explanation: A function is self-inverse if

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

. First, find the inverse of

f(x)

. Let

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

. Swap variables:

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

. Now, solve for

y

x(y-2) = ay+3 \Rightarrow xy - 2x = ay + 3 \Rightarrow xy - ay = 2x+3 \Rightarrow y(x-a) = 2x+3 \Rightarrow y = \frac{2x+3}{x-a}

. So,

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

. For

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

, we need

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

. Comparing rational functions, the coefficients of

x

in the numerators must be equal:

a = 2

. Also, the constant terms in the numerators must be equal:

3 = 3

✓. Finally, the denominators must have the same form, so

-2 = -a

, giving

a = 2

. All conditions confirm

a = 2

Question 3

A function $f$ is such that $f(x+1) = 2f(x) - f(x-1)$ for all $x \in \mathbb{R}$ . Given that $f(1)=5$ and $f(2)=8$ , find $f(4)$ .

11
14 (correct answer)
17
21

Explanation: The recurrence relation can be rearranged as

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

. This shows that the difference between the function values at consecutive integers is constant. This means the sequence of values

f(1), f(2), f(3), ...

forms an arithmetic progression. The common difference is

d = f(2) - f(1) = 8 - 5 = 3

. We can find the next terms:

f(3) = f(2) + d = 8 + 3 = 11

f(4) = f(3) + d = 11 + 3 = 14

. Alternatively, use the given formula directly. For

x=2

f(3) = 2f(2) - f(1) = 2(8) - 5 = 16 - 5 = 11

. For

x=3

f(4) = 2f(3) - f(2) = 2(11) - 8 = 22 - 8 = 14

Question 4

The function $f$ is defined by $f(x) = \ln(3x-5)$ . What is the domain of the inverse function, $f^{-1}$ ?

$x > 0$
$x > 5/3$
$x \in \mathbb{R}$ (correct answer)
$x > \ln(5/3)$

Explanation: The domain of the inverse function

f^{-1}

is equal to the range of the original function

f

. The function is

f(x) = \ln(3x-5)

. First, consider the domain of

f

: we need the argument of the logarithm to be positive, so

3x-5 > 0

, which means

x > 5/3

. Now, consider the range of

f

. As

x

approaches

5/3

from the right,

3x-5

approaches

0^+

, and

\ln(3x-5)

approaches

-\infty

. As

x

approaches

\infty

3x-5

approaches

\infty

, and

\ln(3x-5)

approaches

\infty

. Therefore, the range of

f

(-\infty, \infty)

, which is

\mathbb{R}

. This range is the domain of

f^{-1}

Question 5

Let $f(x) = \sqrt{x+4}$ for $x \ge -4$ . Find the value of $x$ for which $f(x) = f^{-1}(x)$ .

2
$\frac{1+\sqrt{17}}{2}$ (correct answer)
4
$\frac{1+\sqrt{21}}{2}$

Explanation: The intersection of the graph of a function and its inverse often lies on the line

y=x

. So we can solve

f(x)=x

to find the point(s) of intersection.

\sqrt{x+4} = x

. Squaring both sides gives

x+4 = x^2

. Rearranging gives the quadratic equation

x^2 - x - 4 = 0

. Using the quadratic formula,

x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)} = \frac{1 \pm \sqrt{1+16}}{2} = \frac{1 \pm \sqrt{17}}{2}

. Since we squared the equation, we must check for extraneous solutions. The original equation was

\sqrt{x+4}=x

, which implies

x

must be non-negative. Therefore, we must take the positive root:

x = \frac{1+\sqrt{17}}{2}

. We can check that

\sqrt{17}

is slightly more than 4, so

x

is approx

5.1/2 = 2.55

, which is positive.

Question 6

The range of the function $f(x) = a + \ln(x-b)$ is $(-\infty, \infty)$ and its domain is $(2, \infty)$ . The graph of $f(x)$ passes through the point $(3, 5)$ . Find the value of $a$ .

2
3
$e^5$
5 (correct answer)

Explanation: When working with logarithmic functions, you need to understand how transformations affect the domain and range. The function

f(x) = a + \ln(x-b)

is a vertical shift of

\ln(x-b)

a

units. Since the domain is

(2, \infty)

, the expression

(x-b)

must be positive when

x > 2

. This means

x - b > 0

when

x > 2

, so

b = 2

. The natural logarithm

\ln(x-2)

has range

(-\infty, \infty)

when

x > 2

, and adding the constant

a

doesn't change this range, confirming our function has the correct range. Now you can use the given point

(3, 5)

to find

a

. Substituting into the function:

f(3) = a + \ln(3-2) = a + \ln(1) = a + 0 = a = 5

Since

\ln(1) = 0

, we get

a = 5

. Looking at the wrong answers: Choice A (2) likely comes from confusing

a

with

b

, since

b = 2

based on the domain. Choice B (3) might result from mixing up the x-coordinate of the given point with the parameter

a

. Choice C (

e^5

) represents a common error of exponentiating both sides incorrectly or confusing the inverse relationship between logarithms and exponentials. Study tip: With logarithmic transformations, always identify the horizontal shift first using the domain, then use given points to find the vertical shift. The domain tells you what's inside the logarithm, while specific points reveal the vertical translation.

Question 7

Find the domain of the function $f(x) = \sqrt{\ln(x+2)}$ .

$x > -2$
$x \ge -2$
$x > -1$
$x \ge -1$ (correct answer)

Explanation: For the function to be defined, two conditions must be met. First, the argument of the natural logarithm must be positive:

x+2 > 0 \Rightarrow x > -2

. Second, the argument of the square root must be non-negative:

\ln(x+2) \ge 0

. To solve this inequality, we can exponentiate both sides:

e^{\ln(x+2)} \ge e^0

, which simplifies to

x+2 \ge 1

, or

x \ge -1

. We must satisfy both conditions simultaneously:

x > -2

and

x \ge -1

. The intersection of these two conditions is

x \ge -1

. Thus, the domain is

[-1, \infty)

Question 8

Find the range of the function $f(x) = \frac{2x-1}{x+3}$ .

${y \in \mathbb{R} \mid y \ne -3}$
${y \in \mathbb{R} \mid y \ne 1/2}$
${y \in \mathbb{R} \mid y \ne 2}$ (correct answer)
${y \in \mathbb{R} \mid y \ne 3}$

Explanation: One way to find the range of a function is to find the domain of its inverse. Let

y = \frac{2x-1}{x+3}

. To find the inverse, swap

x

and

y

x = \frac{2y-1}{y+3}

. Now solve for

y

x(y+3) = 2y-1 \Rightarrow xy + 3x = 2y-1 \Rightarrow 3x+1 = 2y-xy \Rightarrow 3x+1 = y(2-x) \Rightarrow y = \frac{3x+1}{2-x}

. The domain of this inverse function,

f^{-1}(x)

, is all real numbers except where the denominator is zero, i.e.,

x \ne 2

. Since the domain of

f^{-1}

is the range of

f

, the range of

f(x)

{y \in \mathbb{R} \mid y \ne 2}

. Alternatively, for large

|x|

f(x)

approaches the ratio of the leading coefficients,

y=2/1=2

, which is the horizontal asymptote.

Question 9

Let $f(x) = \sqrt{2x+3} - 1$ . Find the range of $f^{-1}(x)$ .

$y \ge -3/2$ (correct answer)
$y \ge -1$
$y \ge 0$
$y \ge 3/2$

Explanation: The range of the inverse function

f^{-1}(x)

is the domain of the original function

f(x)

. To find the domain of

f(x) = \sqrt{2x+3} - 1

, we must ensure that the expression under the square root is non-negative. So, we require

2x+3 \ge 0

. Solving for

x

gives

2x \ge -3

, which means

x \ge -3/2

. Therefore, the domain of

f(x)

[-3/2, \infty)

. This is the range of

f^{-1}(x)

, so the range is

y \ge -3/2

Question 10

Let $g(x) = \frac{x}{x-1}$ . Which expression is equivalent to $(g \circ g)(x)$ ?

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

Explanation: The expression

(g \circ g)(x)

means

g(g(x))

. To find this, we substitute

g(x)

into

g(x)

itself.

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

. Now substitute the expression for

g(x)

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

. To simplify, first work on the denominator:

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

. Now the full expression becomes

\frac{\frac{x}{x-1}}{\frac{1}{x-1}}

. This simplifies to

\frac{x}{x-1} \cdot \frac{x-1}{1} = x

. This shows that

g(x)

is a self-inverse function.

Question 11

A function $f$ is such that $f(x+1) = 2f(x) - f(x-1)$ for all $x \in \mathbb{R}$ . Given that $f(1)=5$ and $f(2)=8$ , find $f(4)$ .

11
14 (correct answer)
17
21

Explanation: The recurrence relation can be rearranged as

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

. This shows that the difference between the function values at consecutive integers is constant. This means the sequence of values

f(1), f(2), f(3), ...

forms an arithmetic progression. The common difference is

d = f(2) - f(1) = 8 - 5 = 3

. We can find the next terms:

f(3) = f(2) + d = 8 + 3 = 11

f(4) = f(3) + d = 11 + 3 = 14

. Alternatively, use the given formula directly. For

x=2

f(3) = 2f(2) - f(1) = 2(8) - 5 = 16 - 5 = 11

. For

x=3

f(4) = 2f(3) - f(2) = 2(11) - 8 = 22 - 8 = 14

Question 12

The function $f$ is defined by $f(x) = \frac{3}{x+2} - 1$ . What is the range of $f^{-1}(x)$ ?

$y \ne -2$ (correct answer)
$y \ne -1$
$y \ne 1$
$y \ne 3$

Explanation: The range of the inverse function

f^{-1}(x)

is equal to the domain of the original function

f(x)

. The function

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

is a rational function. Its domain is all real numbers except for values of

x

that make the denominator zero. The denominator is

x+2

, which is zero when

x=-2

. Therefore, the domain of

f(x)

\{x \in \mathbb{R} \mid x \ne -2\}

. This domain is the range of

f^{-1}(x)

. So, the range of

f^{-1}(x)

\{y \in \mathbb{R} \mid y \ne -2\}

Question 13

Let $f(x) = \ln(x)$ and $g(x) = 9-x^2$ . Find the domain of the function $h(x) = f(g(x))$ .

$x > 0$
$[-3, 3]$
$(-\infty, -3) \cup (3, \infty)$
$(-3, 3)$ (correct answer)

Explanation: When you encounter a composite function like

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

, finding its domain requires understanding how the functions work together. The domain of a composite function is restricted by both the inner function's domain and the requirement that the inner function's output must be valid input for the outer function. Here,

f(x) = \ln(x)

requires

x > 0

(you can't take the natural logarithm of zero or negative numbers), and

g(x) = 9 - x^2

is defined for all real numbers. Since

h(x) = f(g(x)) = \ln(9 - x^2)

, you need

9 - x^2 > 0

for the logarithm to be defined. Solving

9 - x^2 > 0

: rearrange to get

9 > x^2

, which means

x^2 < 9

. Taking square roots gives

|x| < 3

, or equivalently,

-3 < x < 3

. This is the interval

(-3, 3)

. Answer A)

x > 0

would be correct if you only considered

f(x)

's domain in isolation, ignoring that we're using

g(x)

as the input. Answer B)

[-3, 3]

includes the endpoints where

g(-3) = g(3) = 0

, but

\ln(0)

is undefined. Answer C)

(-\infty, -3) \cup (3, \infty)

represents where

g(x) < 0

, which would make the logarithm undefined. Remember: for composite functions, work from the inside out. The inner function's output becomes the outer function's input, so check that this relationship produces valid results throughout the domain.

Question 14

The function $f$ is defined by $f(x) = x^2 - 4x + 7$ for $x \ge 2$ . Find the expression for $f^{-1}(x)$ .

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

Explanation: To find the inverse function, let

y = f(x)

, swap

x

and

y

, and solve for

y

. Start with

y = x^2 - 4x + 7

. Swap variables:

x = y^2 - 4y + 7

. To solve for

y

, complete the square for the terms involving

y

x = (y^2 - 4y + 4) - 4 + 7

, which simplifies to

x = (y-2)^2 + 3

. Rearrange to isolate

y

x-3 = (y-2)^2

, so

y-2 = \pm\sqrt{x-3}

, and

y = 2 \pm \sqrt{x-3}

. The original function has a domain

x \ge 2

. This becomes the range of the inverse function. So, we must have

f^{-1}(x) \ge 2

. We must choose the positive sign to satisfy this condition. Therefore,

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

Question 15

Find the range of the function $f(x) = \frac{2x-1}{x+3}$ .

${y \in \mathbb{R} \mid y \ne -3}$
${y \in \mathbb{R} \mid y \ne 1/2}$
${y \in \mathbb{R} \mid y \ne 2}$ (correct answer)
${y \in \mathbb{R} \mid y \ne 3}$

Explanation: One way to find the range of a function is to find the domain of its inverse. Let

y = \frac{2x-1}{x+3}

. To find the inverse, swap

x

and

y

x = \frac{2y-1}{y+3}

. Now solve for

y

x(y+3) = 2y-1 \Rightarrow xy + 3x = 2y-1 \Rightarrow 3x+1 = 2y-xy \Rightarrow 3x+1 = y(2-x) \Rightarrow y = \frac{3x+1}{2-x}

. The domain of this inverse function,

f^{-1}(x)

, is all real numbers except where the denominator is zero, i.e.,

x \ne 2

. Since the domain of

f^{-1}

is the range of

f

, the range of

f(x)

{y \in \mathbb{R} \mid y \ne 2}

. Alternatively, for large

|x|

f(x)

approaches the ratio of the leading coefficients,

y=2/1=2

, which is the horizontal asymptote.

Question 16

The function $f(x) = -2(x+3)^2 + 5$ . Which of the following is a possible domain for $f(x)$ on which it has an inverse function?

$[-5, -1]$
$(-\infty, 5]$
$[-3, \infty)$ (correct answer)
$(-\infty, \infty)$

Explanation: A function has an inverse if and only if it is one-to-one. The graph of

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

is a parabola with its vertex at

(-3, 5)

and opening downwards. A parabola is not one-to-one over its entire domain. To make it one-to-one, we must restrict its domain to one side of its axis of symmetry,

x=-3

. The function is increasing on

(-\infty, -3]

and decreasing on

[-3, \infty)

. Any interval that is a subset of one of these two will work. Choice C,

[-3, \infty)

, is the portion of the domain where the function is strictly decreasing, so it is one-to-one on this domain and thus has an inverse.

Question 17

Let $g(x) = \frac{x}{x-1}$ . Which expression is equivalent to $(g \circ g)(x)$ ?

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

Explanation: The expression

(g \circ g)(x)

means

g(g(x))

. To find this, we substitute

g(x)

into

g(x)

itself.

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

. Now substitute the expression for

g(x)

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

. To simplify, first work on the denominator:

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

. Now the full expression becomes

\frac{\frac{x}{x-1}}{\frac{1}{x-1}}

. This simplifies to

\frac{x}{x-1} \cdot \frac{x-1}{1} = x

. This shows that

g(x)

is a self-inverse function.

Question 18

The function $f$ is defined by $f(x) = \ln(3x-5)$ . What is the domain of the inverse function, $f^{-1}$ ?

$x > 0$
$x > 5/3$
$x \in \mathbb{R}$ (correct answer)
$x > \ln(5/3)$

Explanation: The domain of the inverse function

f^{-1}

is equal to the range of the original function

f

. The function is

f(x) = \ln(3x-5)

. First, consider the domain of

f

: we need the argument of the logarithm to be positive, so

3x-5 > 0

, which means

x > 5/3

. Now, consider the range of

f

. As

x

approaches

5/3

from the right,

3x-5

approaches

0^+

, and

\ln(3x-5)

approaches

-\infty

. As

x

approaches

\infty

3x-5

approaches

\infty

, and

\ln(3x-5)

approaches

\infty

. Therefore, the range of

f

(-\infty, \infty)

, which is

\mathbb{R}

. This range is the domain of

f^{-1}

Question 19

Let $f(x) = \sqrt{x+4}$ for $x \ge -4$ . Find the value of $x$ for which $f(x) = f^{-1}(x)$ .

2
$\frac{1+\sqrt{17}}{2}$ (correct answer)
4
$\frac{1+\sqrt{21}}{2}$

Explanation: The intersection of the graph of a function and its inverse often lies on the line

y=x

. So we can solve

f(x)=x

to find the point(s) of intersection.

\sqrt{x+4} = x

. Squaring both sides gives

x+4 = x^2

. Rearranging gives the quadratic equation

x^2 - x - 4 = 0

. Using the quadratic formula,

x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)} = \frac{1 \pm \sqrt{1+16}}{2} = \frac{1 \pm \sqrt{17}}{2}

. Since we squared the equation, we must check for extraneous solutions. The original equation was

\sqrt{x+4}=x

, which implies

x

must be non-negative. Therefore, we must take the positive root:

x = \frac{1+\sqrt{17}}{2}

. We can check that

\sqrt{17}

is slightly more than 4, so

x

is approx

5.1/2 = 2.55

, which is positive.

Question 20

Let $f(x) = x+3$ and $g(x) = \frac{1}{x-1}$ . The function $h$ is defined as $h(x) = (f \circ g)(x)$ . Find the inverse function $h^{-1}(x)$ .

$h^{-1}(x) = \frac{1}{x-3}$
$h^{-1}(x) = \frac{1}{x+3}$
$h^{-1}(x) = 1 + \frac{1}{x+3}$
$h^{-1}(x) = 1 + \frac{1}{x-3}$ (correct answer)

Explanation: When you encounter composite functions and inverse functions together, you need to work systematically: first find the composite function, then find its inverse using the standard algebraic method. Start by finding

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

. Since

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

and

f(x) = x + 3

, you substitute:

h(x) = f\left(\frac{1}{x-1}\right) = \frac{1}{x-1} + 3

. To find

h^{-1}(x)

, use the standard method. Let

y = \frac{1}{x-1} + 3

, then solve for

x

$y - 3 = \frac{1}{x-1}$
$x - 1 = \frac{1}{y-3}$
$x = 1 + \frac{1}{y-3}$

Therefore,

h^{-1}(x) = 1 + \frac{1}{x-3}

. Option A gives

\frac{1}{x-3}

, which misses the "+1" term entirely. Option B uses

\frac{1}{x+3}

, suggesting confusion about which operation to reverse when solving

y - 3 = \frac{1}{x-1}

. Option C gives

1 + \frac{1}{x+3}

, combining both errors: wrong sign in the denominator and missing the correct relationship. You can verify answer D by checking that

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

: substituting

1 + \frac{1}{x-3}

into

h

should return

x

. Study tip: For composite function inverses, always find the composite first, then apply the inverse method. Don't try to find

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

- that's not the same as

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