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📝 Continuity of inverse functions (14 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 14 questions available

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What is Continuity of inverse functions?

Definition:
If a function $f$ is continuous and strictly monotonic (either always increasing or always decreasing) on an interval $I$ , then its inverse function $f^{-1}$ is continuous on the image of $I$ . This means that the inverse of a continuous one-to-one function is also continuous, preserving the property of having no breaks or jumps, which is essential for solving equations and switching between variables.

Example:
$f(x) = e^x$ is continuous and increasing on $\mathbb{R}$ , so its inverse $\ln x$ is continuous on $(0, \infty)$ .
Solution: Since $e^x$ has no breaks, $\ln x$ is also continuous for all $x>0$ .

Reason:
This property allows us to extend continuity to inverse functions, which is crucial for working with logarithms, inverse trig functions, and other inverse relationships in calculus and applied mathematics.

Easy

Medium

Hard

📝 All Continuity of inverse functions MCQs

Q1. Which of the following statements correctly describes Theorem 1.6.2?

A.If f is one‑to‑one and continuous on its domain, then f⁻¹ is continuous on the range of f. ✅

B.If f is continuous, then its inverse exists.

C.If f has a derivative, then its inverse is differentiable.

D.If f is one‑to‑one, then its inverse is always continuous.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The theorem states that continuity of a function together with being one‑to‑one guarantees continuity of its inverse on the function’s range. The other statements either omit the one‑to‑one condition or claim properties not covered by the theorem.

Q2. Suppose f : [0,2]→[0,4] is defined by f(x)=2x. Using Theorem 1.6.2, which of the following must be true about f⁻¹ at y=3?

A.f⁻¹ is discontinuous at y=3.

B.f⁻¹ is continuous at y=3. ✅

C.f⁻¹ does not exist at y=3.

D.f⁻¹ has a jump at y=3.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Since f is linear, one‑to‑one, and continuous on its whole domain, its inverse f⁻¹(y)=y/2 exists for every y in [0,4] and inherits continuity from f by Theorem 1.6.2. Therefore the inverse is continuous at y=3.

Q3. The exponential function f(x)=b^x (b>0, b≠1) is continuous on ℝ. Which statement about its inverse f⁻¹(x)=\log_b x is correct?

A.It is continuous only on (0,∞). ✅

B.It is discontinuous at x=1.

C.It is continuous on ℝ.

D.It has a removable discontinuity at x=0.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The exponential function maps ℝ onto the positive real numbers (0,∞). Its inverse, the logarithm, therefore has domain (0,∞) and, by Theorem 1.6.2, is continuous throughout that domain. It cannot be defined on non‑positive numbers, so options B–D are false.

Q4. Applying Theorem 1.6.2, why is $\sin^{-1}x$ continuous on $[-1,1]$ ?

A.Because

\sin x

is bijective on

[-\pi/2,\pi/2]

and continuous there. ✅

\sin^{-1}x

is defined by a power series.

\sin x

has a derivative on

[-1,1]

\sin^{-1}x

is the composition of continuous functions.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The restricted sine function on $[-\pi/2,\pi/2]$ is one‑to‑one and continuous, with range $[-1,1]$ . By Theorem 1.6.2, the inverse of a continuous one‑to‑one function—here $\sin^{-1}x$ —must be continuous on its domain, which is exactly $[-1,1]$ .

Q5. If a function g has a jump discontinuity at c, what can be inferred about the continuity of its inverse g⁻¹ at the point d=g(c) (assuming g⁻¹ exists)?

A.g⁻¹ must be continuous at d.

B.g⁻¹ must have a jump at d.

C.g⁻¹ cannot be defined at d.

D.No conclusion can be drawn without additional information. ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: A jump in g means the image of a neighborhood of c is split, but the existence of a jump does not uniquely determine the behavior of the inverse at the corresponding y‑value. Additional information about monotonicity or the exact mapping is needed, so no definite conclusion follows.

Q6. Consider the piecewise function k(x)= $\begin{cases}x,&x\le 0\\ x+2,&x>0\end{cases}$ . k is one‑to‑one and has a jump at 0. Determine the continuity of k⁻¹ at y=0.

A.k⁻¹ is continuous at y=0. ✅

B.k⁻¹ has a removable discontinuity at y=0.

C.k⁻¹ has a jump discontinuity at y=0.

D.k⁻¹ is undefined at y=0.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: k⁻¹(y)=y for y≤0 and y−2 for y>2. At y=0, the inverse is defined by the left‑hand formula and there is no approaching sequence from the right because the domain of k⁻¹ does not include (0,2]. Hence the limit from the left equals the function value, giving continuity at y=0.

Q7. Synthesize Theorem 1.6.2 with the composition rule: if f and g are continuous and bijective on their respective domains, what can be said about the continuity of (g ∘ f)⁻¹?

A.It is always discontinuous.

B.It is continuous on the range of g ∘ f. ✅

C.Continuity depends on differentiability of f and g.

D.It is continuous only if f and g are linear.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The composition g ∘ f is itself a continuous bijection; therefore its inverse exists and, by Theorem 1.6.2, is continuous on the range of the composition. No additional smoothness conditions are required.

Q8. If a function f is continuous on its domain but not one‑to‑one, what does Theorem 1.6.2 imply about f⁻¹?

A.f⁻¹ is guaranteed to be continuous.

B.f⁻¹ does not exist as a function. ✅

C.f⁻¹ is continuous at some points.

D.The theorem gives no information about f⁻¹.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The theorem requires the function to be one‑to‑one. When f fails this condition, an inverse function in the usual sense cannot be defined, so the theorem offers no guarantee about continuity.

Q9. Given the restricted cosine function c(x)= $\cos x$ on $[0,\pi]$ , which of the following describes the continuity of its inverse c⁻¹(x)= $\arccos x$ on $[-1,1]$ ?

A.Continuous on (-1,1) only.

B.Discontinuous at x=0.

C.Continuous on the entire interval

[-1,1]

. ✅

D.Has a removable discontinuity at x=1.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: On $[0,\pi]$ the cosine is one‑to‑one and continuous, with range $[-1,1]$ . By Theorem 1.6.2 its inverse, arccos, is continuous on the whole range, i.e., the closed interval $[-1,1]$ .

Q10. Explain why reflecting the graph of a continuous function f across the line y = x guarantees that f⁻¹ has no breaks.

A.Reflection preserves slopes, preventing breaks.

B.Any break in f would correspond to a break in its reflection, contradicting continuity. ✅

C.The line y = x is a symmetry axis that forces continuity.

D.Reflection transforms discontinuities into differentiable points.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A break or hole in the graph of f would appear as a break in the reflected graph because reflection is a one‑to‑one geometric transformation. Since f’s graph is assumed to be without breaks, its mirror image (the graph of f⁻¹) must also be free of breaks, ensuring continuity.

Q11. Let f be continuous and strictly increasing on (a,b). Show that $\displaystyle\lim_{x\to a^{+}}f(x)=L$ exists. Which statement about $\displaystyle\lim_{y\to L^{-}}f^{-1}(y)$ follows from Theorem 1.6.2?

A.The limit does not exist.

B.The limit equals a. ✅

C.The limit equals b.

D.The limit equals L.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Because f is monotone and continuous, its left‑hand limit at the endpoint a exists and equals some finite L. The inverse maps values near L back to x‑values near a, and by Theorem 1.6.2 the inverse is continuous at L, so the limit of f⁻¹ as y→L⁻ is precisely a.

Q12. Using the ε‑δ definition, prove that the inverse of f(x)=x^{3} is continuous at x=0. Which of the following choices for δ in terms of ε correctly guarantees $|\sqrt[3]{y}|<\varepsilon$ whenever $|y|<\delta$ ?

\delta = \varepsilon^{1/3}

\delta = \varepsilon^{3}

✅

\delta = \varepsilon

\delta = \sqrt{\varepsilon}

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: If |y|< $\varepsilon^{3}$ , then taking cube roots gives | $\sqrt[3]{y}$ |< $\varepsilon$ because the cube‑root function is increasing on ℝ. Thus choosing $\delta = \varepsilon^{3}$ satisfies the ε‑δ condition for continuity of the inverse at 0.

Q13. If a function f is continuous on [p,q] and bijective, what can be said about the continuity of its inverse on [f(p),f(q)]?

A.It is continuous only at interior points.

B.It is continuous on the closed interval [f(p),f(q)]. ✅

C.It may have discontinuities at the endpoints.

D.Continuity cannot be determined without differentiability.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A continuous bijection on a closed interval maps the interval onto another closed interval. By Theorem 1.6.2 the inverse function exists and is continuous on the entire image interval, including the endpoints.

Q14. Let f(x)= $\ln x$ (x>0) and g(x)=e^{x}. Both are continuous and bijective on their domains. What is the continuity status of $(g\circ f)^{-1}$ on its domain?

A.Discontinuous at x=1.

B.Continuous on (0,∞). ✅

C.Continuous only on (−∞,0).

D.Not defined because composition is not bijective.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The composition g ∘ f is the identity on (0,∞), which is continuous and bijective. Its inverse is also the identity, hence continuous on the entire domain (0,∞). No break or endpoint issue arises.

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