Q.Every continuous function is differentiable. Examine whether this statement is true.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Continuity Differentiability Relationship
How Continuity and Differentiability Are Related
Two properties describe how "well-behaved" a function is at a point. Continuity means the graph has no break there — you can draw through the point without lifting your pen. Differentiability means the graph is smooth there — it has one definite tangent line, so a well-defined slope f′(a). This concept is about the exact link between the two.
The theorem: If f is differentiable at x=a, then f is continuous at x=a.
Why differentiability forces continuity
If f′(a) exists, then
limx→a(f(x)−f(a))=limx→ax−af(x)−f(a)⋅(x−a)=f′(a)⋅0=0.
So limx→af(x)=f(a), which is exactly continuity at a. A curve that has a tangent cannot also have a jump — a break would send the difference quotient to infinity and the derivative would not exist.
The converse is FALSE
Continuity does not guarantee differentiability. A graph can be unbroken yet still have a sharp corner, and a corner has no single tangent.
The classic counterexample is f(x)=∣x∣ at x=0. It is continuous there (limx→0∣x∣=0=f(0)), but the slope from the left is −1 and from the right is +1. Since these disagree, f′(0) does not exist.
Putting it together
- Differentiable at a ⇒ continuous at a.
- Continuous at a ⇒ differentiable at a.
- Not continuous at a ⇒ not differentiable at a (the contrapositive of the theorem). …
False; ϕ(x)=∣x∣ is a continuous but non-differentiable counterexample.
The modulus function ϕ(x)=∣x∣ is continuous everywhere, including at x=0. But at x=0 the one-sided difference quotients differ:
limh→0+h∣h∣=1,limh→0−h∣h∣=−1, …
The statement is false; the continuous function ϕ(x)=∣x∣ is a counterexample — it is continuous everywhere but not differentiable at x=0.
Step 1 — Note why the claim is tempting.
Many familiar continuous functions are also differentiable everywhere, for example
f(x)=x2,g(x)=ex,h(x)=sinx.
Each is continuous for all x and differentiable for all x, which might suggest that continuity always forces differentiability.
Step 2 — Choose a candidate counterexample.
Consider the modulus function
ϕ(x)=∣x∣={x,−x,x≥0x<0.
It is continuous everywhere, including at x=0, since x→0lim∣x∣=0=ϕ(0).
Step 3 — Test differentiability at x=0.
Examine the difference quotient from each side:
Right derivative: limh→0+h∣0+h∣−∣0∣=limh→0+hh=1,
Left derivative: limh→0−h∣0+h∣−∣0∣=limh→0−h−h=−1. …
Method: Testing a "Does Property A Imply Property B" Claim with a Known Counterexample
Use this whenever asked to examine whether one property (continuity, etc.) forces a stronger related property (differentiability, etc.). The general technique: know the true implication, know why the converse can fail, and keep a standard counterexample ready.
Steps
Step 1: Identify which direction of implication is actually a theorem
For continuity and differentiability, the proven direction is: differentiable at a point ⇒ continuous at that point (provable directly from the definition of the derivative as a limit). Know this direction is solid before questioning the reverse.
Step 2: Recognise the claim under examination is the (unproven) converse
"Every continuous function is differentiable" reverses the arrow of the real theorem. A converse of a true implication is not automatically true — it must be checked independently, usually by trying to find where it fails.
Step 3: Recall or construct a function that is continuous but has a "corner" …
Common Mistakes
Mistake 1: Confusing the claim with the (true) reverse implication
Reasoning "differentiable functions are continuous, so the statement is true" conflates "every continuous function is differentiable" with the actual theorem "every differentiable function is continuous." Why it's wrong: these are different statements; only the second is a proven theorem, and confusing them leads to affirming a false converse. Correct approach: always identify precisely which direction of implication is being asked about before deciding whether it's provably true or needs a counterexample.
Mistake 2: Sign errors in the one-sided derivative limits …
- CBSE 2026Set ANNUAL1 markQ.Write True or False: Every differentiable function is continuous, but the converse is not true.
›Reveal solutionSolution
True - differentiable ⇒ continuous, converse false.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Assertion (A): Every differentiable function is continuous. Reason (R): Every continuous function is differentiable. Choose the correct option:(a) Both A and R are correct and R is the correct explanation of A.(b) Both A and R are correct but R is not the correct explanation of A.(c) A is correct but R is incorrect.(d) Both A and R are incorrect.
›Reveal solutionSolution
Differentiability implies continuity, but the converse is false.
A is TRUE: Every differentiable function is continuous — this is a standard theorem.
…
- CBSE 2023Set 65/3/11 markMCQQ.The value of k for which function f(x)={x2,kx,x≥0x<0 is differentiable at x=0 is :(a) 1(b) 2(c) any real number(d) 0
›Reveal solutionSolution
A piecewise function is differentiable at a point only if it is continuous there and the left and right derivatives match. For f(x) at x=0, continuity forces k to be anything, but matching derivatives forces k=0.
Differentiability is a stronger condition than continuity. For a function to be differentiable at a point, two things must happen: the function must be continuous there, and the derivative must exist (meaning the left-hand and right-hand derivatives must be equal).
Let's check both conditions systematically for f(x) at x=0.
Checking Continuity at x=0
For continuity at x=0, we need:
limx→0−f(x)=limx→0+f(x)=f(0)
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From the right: As x→0+, we use f(x)=x2, so limx→0+f(x)=02=0.
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From the left: As x→0−, we use f(x)=kx, so limx→0−f(x)=k⋅0=0.
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At the point: f(0)=02=0 (since 0≥0, we use the first piece).
All three equal 0 regardless of k, so f is continuous at x=0 for any value of k.
Checking Differentiability at x=0
Now we compute the left-hand derivative (LHD) and right-hand derivative (RHD) using the definition:
f′(0)=limh→0hf(0+h)−f(0)=limh→0hf(h)
Right-hand derivative (h→0+): …
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- CBSE 2020Set HE8231 markQ.Fill in the blank: Every differentiable function is ______.
›Reveal solutionSolution
Every differentiable function is continuous.
If a function f is differentiable at a point x=a, then
limh→0hf(a+h)−f(a)=f′(a) exists (finite).
Writing f(a+h)−f(a)=hf(a+h)−f(a)⋅h and taking the limit as h→0:
limh→0[f(a+h)−f(a)]=f′(a)⋅0=0,
which means h→0limf(a+h)=f(a) — exactly the condition for continuity at x=a.
…
- CBSE 2019Set ANNUAL1 markQ.If f′(2+)=0 and f′(2−)=0, then is f(x) continuous at x=2?
›Reveal solutionSolution
Existence of a finite one-sided derivative at a point forces one-sided continuity there; since both f′(2+) and f′(2−) exist (and equal 0), f is continuous at x=2.
By definition, f′(2+)=h→0+limhf(2+h)−f(2).
For this limit to exist and be finite (here, equal to 0), the numerator f(2+h)−f(2) must itself tend to 0 as h→0+ — because f(2+h)−f(2)=h⋅hf(2+h)−f(2)→0⋅0=0.
So h→0+limf(2+h)=f(2), i.e. f is right-continuous at x=2.
…
- CBSE 2018Set ANNUAL1 markMCQQ.If f(x) is differentiable at x = a, which of the following statement may be false ?(a) f(x) is continuous at x = a(b) lim x→a f(x) exist(c) lim h→0⁻ [f(a+h) − f(a)]/h = lim h→0⁺ [f(a+h) − f(a)]/h(d) The second derivative of f(x) i.e. f″(x) exist at x = a
›Reveal solutionSolution
Differentiability at a point guarantees continuity and the existence of f′(a), but never guarantees f″(a) exists.
If f is differentiable at x=a, then by definition the limit f′(a)=limh→0hf(a+h)−f(a) exists, which forces:
- (a) f continuous at a — always true (differentiability ⇒ continuity).
- (b) limx→af(x) exists — always true, again because differentiability implies continuity.
- (c) limh→0−hf(a+h)−f(a)=limh→0+hf(a+h)−f(a) — always true, since a two-sided derivative existing means the left-hand and right-hand derivatives are equal. …
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