📝 Inverse functions definition and properties (17 MCQs)

📖 From Calculus • 1. Basics before calculus • 17 questions available

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What is Inverse functions definition and properties?

Definition:
An inverse function, denoted $f^{-1}$ , reverses the effect of a function $f$ , satisfying $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$ , and $f(f^{-1}(y)) = y$ for all $y$ in the range, provided $f$ is one-to-one.

Example:
For $f(x) = 2x+3$ , inverse is $f^{-1}(y) = (y-3)/2$ , since $f^{-1}(f(x)) = (2x+3-3)/2 = x$ .

Reason:
Inverses allow us to undo" operations, essential in solving equations and in many applications like encryption and decoding."

Easy

Medium

Hard

📝 All Inverse functions definition and properties MCQs

Q1. If $h$ is the inverse of $f$ and $k$ is the inverse of $h$ , what is the relationship between $k$ and $f$ ?

k = f^{-1}

k = f

✅

k = f^{-1}

k

is unrelated to

f

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Taking the inverse of a function twice restores the original function: $(f^{-1})^{-1}=f$ . Since $h=f^{-1}$ and $k=h^{-1}$ , it follows that $k=(f^{-1})^{-1}=f$ . Thus $k$ and $f$ are the same function.

Q2. Suppose a function $p$ satisfies $p(p^{-1}(x))=x$ for all $x$ in its range, but $p^{-1}(p(x))=x$ fails for some $x$ . Which statement must be false?

p

is one‑to‑one

p

is onto

p

has an inverse

p

is bijective ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: If the second cancellation fails, the function cannot be onto as well as one‑to‑one; bijectivity requires both properties. Therefore the claim that $p$ is bijective must be false, even though it may still be one‑to‑one.

Q3. For $f(x)=x^{3}+1$ and its inverse $f^{-1}(x)=\sqrt[3]{\,x-1\,}$ , which description of their domains is correct?

A.Both have domain

\mathbb{R}

B.Domain of

f

\mathbb{R}

but range of

f^{-1}

[1,\infty)

✅

C.Domain of

f^{-1}

[1,\infty)

only

D.Both are defined only for

x\ge 1

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The cubic function $x^{3}+1$ is defined for every real $x$ , giving a range of all real numbers. Its inverse therefore has domain equal to that range, i.e., $\mathbb{R}$ . Both functions are defined for all real inputs.

Q4. What is $(f^{-1}\circ f)(2)$ when $f(x)=x^{3}+1$ ?

2

✅

8

\sqrt[3]{9}

1

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The composition $f^{-1}(f(2))$ first evaluates $f(2)=2^{3}+1=9$ and then applies $f^{-1}$ to 9, yielding $\sqrt[3]{9-1}=2$ . Thus the overall result is the original input, $2$ .

Q5. Which geometric description correctly relates the graph of a function to the graph of its inverse?

A.They are reflections of each other across the line

y=x

✅

B.They are reflections across the

x

-axis

C.They are reflections across the

y

-axis

D.There is no systematic relationship

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: By definition, swapping the roles of $x$ and $y$ reflects the set of points $(x,y)$ across the line $y=x$ . Hence the graph of a function and that of its inverse are mirror images about this line.

Q6. What does the notation $f^{-1}(x)$ represent?

A.The reciprocal of

f(x)

B.The inverse function of

f

evaluated at

x

✅

C.The derivative of

f

x

D.A negative exponent applied to

f

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The superscript $-1$ on a function symbol denotes the inverse function, not a reciprocal or derivative. Thus $f^{-1}(x)$ means “apply the inverse of $f$ to the input $x$ .”

Q7. Why does interchanging $x$ and $y$ in the equation $y = x^{3}+1$ lead to the inverse function?

A.Solving for

x

gives

x = \sqrt[3]{\,y-1\,}

, which defines

f^{-1}

✅

B.Taking the reciprocal of both sides yields the inverse

C.Differentiating both sides produces the inverse

D.Integrating both sides yields the inverse

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Swapping the variables treats the original output as the new input. Solving the resulting equation for the new dependent variable isolates $x$ as $\sqrt[3]{\,y-1\,}$ , which is precisely the expression for the inverse function.

Q8. For a strictly increasing function $f$ , which property guarantees that $f$ possesses an inverse?

A.Continuity on its domain

B.Being one‑to‑one (injective) ✅

C.Having a horizontal asymptote

D.Differentiability everywhere

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A strictly increasing function never takes the same value twice, making it injective. Injectivity is the essential condition for the existence of an inverse; continuity or differentiability are not required, though they often accompany monotonicity.

Q9. What is the inverse of $f(x)=2x+3$ ?

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

✅

2x-3

\displaystyle\frac{x+3}{2}

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

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Set $y=2x+3$ , solve for $x$ : $x=\frac{y-3}{2}$ . Replacing $y$ with the input variable yields the inverse function $f^{-1}(x)=\frac{x-3}{2}$ .

Q10. If $f^{-1}(x)=\sqrt[3]{\,x-1\,}$ , what is $f(x)$ ?

x^{3}+1

✅

\sqrt[3]{x}+1

(x+1)^{3}

x^{1/3}+1

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The inverse relationship means that solving $y=\sqrt[3]{\,x-1\,}$ for $x$ gives $x=y^{3}+1$ . Replacing $y$ with the original variable $x$ produces $f(x)=x^{3}+1$ .

Q11. For $f(x)=x^{3}+1$ , what is $(f\circ f^{-1})(5)$ ?

5

✅

125+1

\sqrt[3]{5-1}+1

D.Undefined

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: First apply $f^{-1}$ to 5: $f^{-1}(5)=\sqrt[3]{5-1}= \sqrt[3]{4}$ . Then apply $f$ : $f(\sqrt[3]{4})=(\sqrt[3]{4})^{3}+1=4+1=5$ . The composition returns the original argument, 5.

Q12. True or false: Every function has an inverse.

A.true

B.false ✅

C.It depends on the function

D.Cannot be determined

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A function must be one‑to‑one (injective) to possess an inverse. Many functions fail this condition, such as $f(x)=x^{2}$ on $\mathbb{R}$ . Hence the statement is false.

Q13. Which statement correctly describes the relationship between the domains of a function and its inverse?

A.The domain of

f

equals the range of

f^{-1}

✅

B.The domain of

f

equals the domain of

f^{-1}

C.The range of

f

equals the range of

f^{-1}

D.There is no systematic relationship

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: By definition, the set of outputs (range) of $f$ becomes the set of permissible inputs (domain) for $f^{-1}$ . Consequently, the domain of $f$ is precisely the range of its inverse, and vice versa.

Q14. Suppose $f$ is defined on $\mathbb{R}$ and $f^{-1}$ exists. If $f(4)=10$ , what is $(f^{-1})'(10)$ in terms of $f'(4)$ ?

\displaystyle\frac{1}{f&#039;(4)}

✅

f&#039;(4)

-f&#039;(4)

\displaystyle\frac{1}{[f&#039;(4)]^{2}}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The derivative of an inverse satisfies $(f^{-1})'(y)=\frac{1}{f'(x)}$ where $y=f(x)$ . Here $y=10$ and $x=4$ , so $(f^{-1})'(10)=1/f'(4)$ .

Q15. Given $f(x)=x^{3}+1$ , for which $x$ does $(f^{-1})'(x)$ exist?

A.All real numbers

x>1

x\neq 1

✅

D.No values

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The inverse derivative exists wherever $f'(x)\neq0$ . Since $f'(x)=3x^{2}$ , it vanishes only at $x=0$ , corresponding to $y=f(0)=1$ . Thus $(f^{-1})'(x)$ is defined for every $x$ except $x=1$ .

Q16. When graphing $f(x)=x^{3}+1$ and its inverse together, which transformation maps the graph of $f$ onto the graph of its inverse?

A.Reflection across the line

y=x

✅

B.Rotation by

90^\circ

about the origin

C.Horizontal translation by 1 unit

D.Uniform scaling by a factor of 2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Swapping the roles of $x$ and $y$ corresponds to reflecting each point $(x,y)$ across the line $y=x$ . This operation converts the graph of a function into the graph of its inverse.

Q17. If a function satisfies $f^{-1}(x)=\frac{1}{x}$ , which of the following could be $f(x)$ ?

x

\frac{1}{x}

✅

-x

x^{2}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The function $g(x)=1/x$ is its own inverse because solving $y=1/x$ for $x$ yields $x=1/y$ , which is the same expression. Therefore the original function $f$ that has this inverse must also be $f(x)=1/x$ .

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