📝 How to find inverse of a function (15 MCQs)

📖 From Calculus • 1. Basics before calculus • 15 questions available

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What is How to find inverse of a function?

Definition:
To find the inverse of a function $y = f(x)$ , swap $x$ and $y$ , then solve for $y$ in terms of $x$ , resulting in $y = f^{-1}(x)$ , provided the original function is one-to-one.

Example:
For $f(x) = \sqrt{x-1}$ with domain $x \geq 1$ , set $y = \sqrt{x-1}$ , swap to $x = \sqrt{y-1}$ , solve: $x^2 = y-1 \Rightarrow y = x^2+1$ , so $f^{-1}(x) = x^2+1$ with domain $x \geq 0$ .

Reason:
This algebraic method provides a systematic way to reverse mappings, useful for solving equations and understanding symmetry.

Easy

Medium

Hard

📝 All How to find inverse of a function MCQs

Q1. What is the first step in the standard procedure for finding the inverse of a function?

A.Write the equation

y = f(x)

. ✅

B.Solve the equation for

x

in terms of

y

C.Interchange

x

and

y

in the resulting expression.

D.State the domain of the inverse function.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The procedure begins by expressing the function as an equation with $y$ on the left side, $y = f(x)$ . This step makes the relationship explicit and prepares the equation for subsequent algebraic manipulation. Without this initial statement, the later steps of solving for $x$ and swapping variables cannot be carried out correctly.

Q2. Given $f(x)=x^{3}+1$ and its inverse $f^{-1}(x)=\sqrt[3]{x-1}$ , which statement must be true for all $x$ in the domain of $f$ ?

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

for all

x

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

only for positive

x

C.Both compositions

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

and

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

equal

x

. ✅

D.The inverse does not satisfy any composition identity.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: By definition of inverse functions, applying the original function to its inverse and vice‑versa returns the original input. Therefore both compositions $f(f^{-1}(x))$ and $f^{-1}(f(x))$ must equal $x$ for every $x$ in the appropriate domains. This property guarantees that the two functions truly undo each other.

Q3. Which of the following describes the method that produces an inverse formula with $x$ as the independent variable without a final swapping step?

A.Solve

y = f(x)

for

x

and then replace

y

x

B.Reverse the roles of

x

and

y

at the outset and solve

x = f(y)

for

y

. ✅

C.Differentiate

f

and integrate the result.

D.Apply the theorem directly without solving.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If we start by interchanging the symbols, we write $x = f(y)$ and then solve directly for $y$ . The solution yields $y$ expressed in terms of the original $x$ , so the independent variable is already $x$ . This avoids a later swapping step, producing the inverse function in the desired form.

Q4. For $f(x)=\sqrt{3x-2}$ , the inverse is $f^{-1}(x)=\frac{1}{3}(x^{2}+2)$ . What is the correct domain of $f^{-1}$ and why does it differ from the natural domain of the formula?

A.All real numbers, because the formula is defined everywhere.

x\le 0

, because the square root restricts the range.

x\ge 0

, because the range of

f

[0,\infty)

. ✅

D.No domain restriction, the inverse is defined for any

x

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The original function $f$ yields only non‑negative outputs, so its range is $[0,\infty)$ . That range becomes the domain of the inverse function. Although the algebraic expression $\frac{1}{3}(x^{2}+2)$ is defined for all real $x$ , the inverse is only meaningful for $x$ values that actually arise from $f$ , namely $x\ge 0$ .

Q5. If a function is not one‑to‑one on its whole domain, can the procedure described in the theorem still produce an inverse function?

A.Yes, the procedure always yields an inverse.

B.No, an inverse does not exist unless the function is restricted to a one‑to‑one subset. ✅

C.Only if the function is linear.

D.The procedure works but gives a multivalued relation.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: An inverse function requires each output to correspond to exactly one input, i.e., the function must be one‑to‑one. When the original function fails this condition, no single‑valued inverse exists on the full domain. The theorem’s procedure can still be applied after restricting the domain to a region where the function becomes one‑to‑one, thereby producing a valid inverse.

Q6. Given $f(x)=2x+5$ and $g(x)=\frac{x-5}{2}$ , which statement correctly identifies the inverse relationship?

g

is the inverse of

f

. ✅

f

is the inverse of

g

C.Neither is an inverse of the other.

D.Both are inverses of each other.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Substituting $g$ into $f$ gives $f(g(x))=2\left(\frac{x-5}{2}\right)+5=x$ . Similarly, $g(f(x))=\frac{(2x+5)-5}{2}=x$ . Both compositions return the original input, confirming that $g$ undoes $f$ and vice versa, so $g$ is indeed the inverse of $f$ .

Q7. Find the inverse of $f(x)=\frac{x-1}{x+2}$ and select the correct expression.

\frac{1-2x}{x+1}

\frac{2x-1}{x-1}

\frac{1+2x}{1-x}

✅

\frac{x+1}{1-2x}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Starting with $y=\frac{x-1}{x+2}$ , cross‑multiply to obtain $y(x+2)=x-1$ . Rearranging gives $yx+2y=x-1$ → $x(y-1)=-(1+2y)$ . Solving for $x$ yields $x=\frac{1+2y}{1-y}$ . Replacing $y$ with $x$ gives the inverse $f^{-1}(x)=\frac{1+2x}{1-x}$ .

Q8. Suppose $h$ is strictly increasing on its domain and has an inverse $h^{-1}$ . Which of the following must be true about $h^{-1}$ ?

A.It is strictly decreasing. ✅

B.It is constant.

C.It is also strictly increasing.

D.It has no monotonic behavior.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A strictly increasing function preserves order: if $a<b$ then $h(a)<h(b)$ . Applying the inverse to both sides yields $h^{-1}(h(a))=a$ and $h^{-1}(h(b))=b$ , so the order of inputs is maintained. Consequently, the inverse function must also be strictly increasing, mirroring the monotonicity of the original function.

Q9. In the inverse‑finding procedure, which step guarantees that the final formula uses $x$ as the independent variable?

A.Writing

y = f(x)

B.Interchanging

x

and

y

after solving for

x

. ✅

C.Checking the domain of the original function.

D.Differentiating the obtained expression.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: After solving the original equation for $x$ in terms of $y$ , the resulting expression still treats $y$ as the independent variable. Swapping the symbols $x$ and $y$ converts the dependent variable to the independent position, producing a formula $y = f^{-1}(x)$ where $x$ is now the input variable.

Q10. Why is the domain of $f^{-1}$ equal to the range of $f$ in the inverse‑function theorem?

A.Because the output of

f

becomes the input of

f^{-1}

by definition. ✅

B.Because both functions share the same formula.

C.Because the domain of

f

is always

[0,\infty)

D.Because the inverse swaps the roles of coefficients.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: By definition, an inverse function reverses the mapping of the original function: each output value of $f$ serves as an input for $f^{-1}$ . Therefore, the set of all possible outputs of $f$ (its range) precisely forms the set of allowable inputs for $f^{-1}$ , establishing equality of the two sets.

Q11. Let $f(x)=x^{2}$ with domain $x\ge 0$ . What is $f^{-1}(x)$ ?

-\sqrt{x}

2\sqrt{x}

x^{2}

\sqrt{x}

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: When the domain of $f$ is restricted to non‑negative numbers, the function becomes one‑to‑one, allowing an inverse. Solving $y = x^{2}$ for $x$ yields $x = \sqrt{y}$ . Replacing $y$ with the independent variable gives $f^{-1}(x)=\sqrt{x}$ , which is defined for $x\ge 0$ .

Q12. For $f(x)=\frac{3x+1}{2x-5}$ , the inverse is $f^{-1}(x)=\frac{5x+1}{2x-3}$ . What is the domain of $f^{-1}$ ?

A.All real numbers.

B.All real numbers except

x=5/2

C.All real numbers except

x=3/2

. ✅

D.Only non‑negative numbers.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The inverse formula contains the denominator $2x-3$ ; it is undefined when $2x-3=0$ , i.e., $x=3/2$ . This value also corresponds to the horizontal asymptote of the original function, which is excluded from its range. Consequently, the domain of the inverse consists of all real numbers except $x=3/2$ .

Q13. Which statement best describes the relationship between a function and its inverse?

A.They have identical graphs.

B.They exchange the roles of inputs and outputs. ✅

C.One is always a translation of the other.

D.The inverse is the derivative of the original function.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: An inverse function reverses the mapping of the original: every ordered pair $(a,b)$ in the function becomes $(b,a)$ in the inverse. This exchange of inputs and outputs is the defining characteristic of inverse functions, and it is reflected graphically by reflecting the original graph across the line $y=x$ .

Q14. If after solving $y = f(x)$ for $x$ you forget to interchange $x$ and $y$ , what is the most likely result?

A.The expression represents

f^{-1}

with the wrong independent variable, causing domain confusion. ✅

B.You obtain the original function again.

C.The formula becomes undefined for all

x

D.The inverse will be identical to the original function.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Solving for $x$ yields a formula $x = g(y)$ where $y$ is still the independent variable. Without swapping the symbols, the expression cannot be used as a function of $x$ ; its domain will correspond to the range of the original function, leading to mismatched variables and potential confusion when applying the inverse.

Q15. If $f$ and $g$ are inverse functions, which identity must hold for every $x$ in the domain of $f$ ?

f(x)+g(x)=0

f(g(x))=x

g(f(x))=x

D.Both B and C. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: By definition of inverses, composing one function with the other in either order returns the original input: $f(g(x))=x$ and $g(f(x))=x$ . Both identities are required for the functions to truly be inverses of each other, ensuring that each undoes the effect of the other.

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