📝 Differentiability of a function (18 MCQs)

📖 From Calculus • 3. The Derivation • 18 questions available

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What is Differentiability of a function?

Definition:
A function is differentiable at a point if its derivative exists there, meaning the limit defining the derivative converges to a finite value, implying the graph has no sharp corners, cusps, or vertical tangents at that specific location.

Example:
$f(x) = |x|$ is not differentiable at $x=0$ because left derivative is $-1$ and right derivative is $1$ , so they do not match.

Reason:
Checking differentiability ensures functions are smooth enough for calculus operations, preventing errors in optimization and integration where undefined derivatives would invalidate standard techniques and results.

Easy

Medium

Hard

📝 All Differentiability of a function MCQs

Q1. For f(x)=|x|, which statement about differentiability at x=0 is correct?

A.Derivative exists and equals 0

B.Derivative does not exist ✅

C.Derivative exists and equals 1

D.Derivative exists and equals -1

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The left‑hand limit of the difference quotient gives -1 while the right‑hand limit gives +1, so the two one‑sided limits differ. Because the two‑sided limit fails to exist, the derivative at x=0 does not exist.

Q2. If f is differentiable at x₀, which statement must be true?

A.f is continuous at x₀ ✅

B.f has a vertical tangent at x₀

C.f is not continuous at x₀

D.f has a corner at x₀

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Differentiability implies the limit of the difference quotient exists, which forces the function to be continuous at that point. Continuity is therefore a necessary condition, whereas the other options describe situations where differentiability fails.

Q3. Suppose g is continuous at a and its left‑hand derivative is –2 while its right‑hand derivative is 3. What can be concluded about differentiability at a?

A.g is differentiable at a

B.g is not differentiable at a ✅

C.The derivative at a equals 0

D.The derivative is undefined but g has a vertical tangent

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Differentiability requires the two one‑sided limits of the difference quotient to coincide. Here the left‑hand limit equals –2 and the right‑hand limit equals 3, so the two‑sided limit does not exist, meaning g is not differentiable at a.

Q4. Compare the differentiability of h(x)=x^{1/3} and k(x)=x^{2/3} at x=0. Which statement is true?

A.Both are differentiable at 0

B.h is differentiable, k is not

C.k is differentiable, h is not

D.Neither is differentiable at 0 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: For h, the difference quotient becomes h^{-2/3} which diverges as h→0; for k it becomes h^{-1/3}, also diverging. Since the limits are infinite, neither function possesses a finite derivative at the origin.

Q5. If a function f has a vertical tangent at x₀, which describes the behavior of \ $\\lim_{h\\to0}\\frac{f(x₀+h)-f(x₀)}{h}\$ ?

A.The limit equals 0

B.The limit equals +\\infty

C.The limit does not exist (unbounded) ✅

D.The limit equals a finite non‑zero number

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A vertical tangent means the slope of secant lines grows without bound as the point approaches x₀ from either side. Hence the difference quotient becomes unbounded, so the limit fails to exist in the finite sense.

Q6. Consider p(x)=x^{2}\\sin(1/x) for x\\neq0 and p(0)=0. Determine differentiability at 0 and justify using the definition.

A.Differentiable with derivative 0 ✅

B.Not differentiable

C.Differentiable but derivative does not exist

D.Not differentiable but continuous

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using the definition, \ $\\frac{p(h)-p(0)}{h}=h\\sin(1/h)\$ . Since \ $|\\sin|\\le1\$ , the absolute value is bounded by |h|, which tends to 0 as h→0. Thus the limit exists and equals 0, so p is differentiable at the origin.

Q7. Let f be differentiable on (‑2,2) with f(‑1)=3 and f(1)=‑1. Which statement must be true?

A.There exists c in (‑1,1) with f′(c)=‑2

B.There exists c with f′(c)=2

C.There exists c with f′(c)=0 ✅

D.No conclusion about f′ can be drawn

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: By the Mean Value Theorem, there is some c between –1 and 1 such that f′(c)=\ $\\frac{f(1)-f(-1)}{1-(-1)}\$ =\ $\\frac{-1-3}{2}\$ =‑2. Hence the existence of such a c is guaranteed.

Q8. Suppose q(x)=x^{2} for x\\le0 and q(x)=x for x>0. At x=0, which statement about differentiability is correct?

A.q is differentiable at 0 with derivative 0

B.q is not differentiable because left derivative \\neq\\ right derivative ✅

C.q is differentiable at 0 with derivative 1

D.q is not differentiable because it is discontinuous at 0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The left‑hand derivative at 0 is \ $\\lim_{h\\to0^-}\\frac{h^{2}-0}{h}=0\$ , while the right‑hand derivative is \ $\\lim_{h\\to0^+}\\frac{h-0}{h}=1\$ . Because the one‑sided derivatives differ, the overall derivative does not exist.

Q9. Which statement best explains why continuity is necessary but not sufficient for differentiability?

A.Continuity guarantees a tangent line

B.Discontinuities cause undefined derivative

C.A function can be continuous but have a corner ✅

D.Continuity implies the derivative exists

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A function may be continuous everywhere yet possess a corner (as with \ $|x|\$ at 0). At a corner the left‑ and right‑hand slopes differ, so the derivative fails to exist despite the function being continuous.

Q10. What is the geometric interpretation of the derivative at a point?

A.Slope of a secant line

B.Slope of the tangent line ✅

C.Curvature of the graph

D.Area under the curve

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The derivative at a point gives the instantaneous rate of change, which corresponds to the slope of the line that just touches the graph at that point—the tangent line.

Q11. Given f(x)=|x|^{3}, determine differentiability at x=0.

A.Differentiable with derivative 0 ✅

B.Not differentiable

C.Derivative does not exist

D.Differentiable with derivative 3

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For x>0, f(x)=x^{3}; for x<0, f(x)=‑x^{3}. The left‑hand derivative is \ $-3x^{2}\$ →0 and the right‑hand derivative is \ $3x^{2}\$ →0 as x→0, so both one‑sided limits agree and equal 0.

Q12. If a function r has derivative r′(x)=0 for all x in an interval, what can be inferred about r on that interval?

A.r is constant on the interval ✅

B.r is linear on the interval

C.r is increasing on the interval

D.r attains a maximum on the interval

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A zero derivative everywhere on an interval means the function does not change value there; by the Mean Value Theorem, r must be constant throughout the interval.

Q13. Consider s(x)=x\\sin(1/x) for x\\neq0 and s(0)=0. Which statement about differentiability at 0 is true?

A.Differentiable with derivative 0

B.Not differentiable ✅

C.Differentiable but derivative does not exist

D.Not differentiable but continuous

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The difference quotient \ $\\frac{s(h)-0}{h}=\\sin(1/h)\$ oscillates between –1 and 1 as h→0, so the limit does not exist. Hence s is not differentiable at the origin, although it remains continuous.

Q14. Let f(x)=x^{2} for rational x and f(x)=0 for irrational x. Determine differentiability at x=0.

A.Differentiable with derivative 0 ✅

B.Not differentiable

C.Differentiable but derivative does not exist

D.Continuous but not differentiable

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Both rational and irrational sequences approaching 0 give function values tending to 0, so f is continuous at 0. The difference quotient equals h for rational h and 0 for irrational h; both tend to 0, yielding a derivative of 0.

Q15. Which is a necessary condition for a function to have a vertical tangent at x₀?

A.Derivative is infinite at x₀ ✅

B.Function is discontinuous at x₀

C.Left‑hand and right‑hand derivatives are equal and finite

D.Function is monotonic near x₀

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A vertical tangent occurs when the slope of the secant lines grows without bound, which mathematically means the derivative approaches ±∞. Continuity or monotonicity are not required, but an unbounded derivative is necessary.

Q16. If a function is not differentiable at x₀, which could be a reason?

A.Corner point

B.Vertical tangent

C.Cusp

D.All of the above ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: All three listed phenomena—corners, vertical tangents, and cusps—lead to failure of the limit defining the derivative, so any of them can explain nondifferentiability at the point.

Q17. Suppose f is differentiable everywhere and satisfies f′(x)=f(x) for all x. Which function could be f?

A.f(x)=0

B.f(x)=e^{x} ✅

C.f(x)=\\sin x

D.f(x)=x^{2}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The differential equation f′=f has the general solution f(x)=Ce^{x}. Choosing C=1 gives f(x)=e^{x}, which indeed satisfies the condition. The other options do not satisfy f′=f.

Q18. A function g is defined by g(x)=\\begin{cases}x^{2}\\sin(1/x) & x\\neq0\\\\0 & x=0\\end{cases}. Which description of the graph near the origin regarding differentiability and tangency is correct?

A.Differentiable with a horizontal tangent ✅

B.Not differentiable, cusp

C.Differentiable with a vertical tangent

D.Not differentiable, corner

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The difference quotient \ $\\frac{g(h)-0}{h}=h\\sin(1/h)\$ tends to 0 as h→0, confirming differentiability with derivative 0. Consequently the tangent line at the origin is horizontal, and the graph flattens out despite oscillations.

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