Concept understanding — Differentiability and Continuity
A function can be continuous at a point yet still fail to have a derivative there — differentiability is a strictly stronger requirement than continuity, never the other way around.
Three ways a derivative can fail to exist at x0:
A corner or cusp (the graph comes to a sharp point ∨ or ∧). Example: f(x)=∣x−2∣ at x=2 — continuous there, but f′(2−)=−1=1=f′(2+), so f′(2) does not exist.
A vertical tangent. Example: f(x)=x1/3 at x=0 — continuous, but f′(0)=limx→0x−2/3=+∞, not a finite number.
A discontinuity. Example: f(x)=⌊x⌋ at any integer n — not even continuous there, so certainly not differentiable; or a jump function like f(x)=x (x≤0), f(x)=x+1 (x>0), where f′(0+) fails to exist because the right-hand difference quotient blows up as the jump is approached.
These three cases are exhaustive: a function fails to be differentiable at a point of its domain precisely when one of them holds. In short, discontinuity always forces non-differentiability — but continuity alone never guarantees differentiability, as cases (1) and (2) show.
The one implication that always holds:
Theorem 10.1. If f is differentiable at x0, then f is continuous at x0.
Proof. Since f′(x0)=limΔx→0Δxf(x0+Δx)−f(x0) exists, write f(x0+Δx)−f(x0)=Δxf(x0+Δx)−f(x0)×Δx. Taking the limit of both sides as Δx→0 (product of limits), limΔx→0[f(x0+Δx)−f(x0)]=f′(x0)×0=0, i.e. limΔx→0f(x0+Δx)=f(x0) — which is exactly continuity of f at x0. ■
Watch out
The converse of Theorem 10.1 is false: ∣x−2∣ and x1/3 are both continuous at the flagged point but not differentiable there. "Continuous" is a necessary but never a sufficient condition for "differentiable".
Compute limh→0−hf(1+h)−f(1) and limh→0+hf(1+h)−f(1) separately for each function.
✓Final answer
(i) LHD=−1, RHD=1 — not differentiable at x=1.
(ii) only the LHD exists and →−∞ (vertical tangent); the RHD is undefined since f isn't real for x>1 — not differentiable at x=1.
(iii) LHD=1, RHD=2 — not differentiable at x=1.
Step 1. For each function, f′(1−)=limh→0−hf(1+h)−f(1) and f′(1+)=limh→0+hf(1+h)−f(1); f is differentiable at x=1 only if both exist and are equal.
Step 2. (i)f(x)=∣x−1∣, so f(1)=0 and f(1+h)=∣h∣.
hf(1+h)−f(1)=h∣h∣.
For h→0−, h<0⇒∣h∣=−h, so the quotient =h−h=−1. Thus f′(1−)=−1.
For h→0+, h>0⇒∣h∣=h, so the quotient =hh=1. Thus f′(1+)=1.
Since −1=1, f is not differentiable at x=1.
Step 3. (ii)f(x)=1−x2, domain [−1,1]. f(1)=0, and for x>1 the function is not real, so f′(1+) has no meaning — only the left derivative can be tested.
f(1+h)=1−(1+h)2=−2h−h2=−h(2+h).
For h→0−, write h=−∣h∣ so −h=∣h∣>0 and −h(2+h)=∣h∣(2+h)>0 (valid, matches the domain). Then
As h→0−, 2+h→2 while ∣h∣→0+, so the quotient →−∞.
So f′(1−) does not exist as a finite number (the curve has a vertical tangent at x=1), and f′(1+) is not even defined since f isn't real for x>1. Hence f is not differentiable at x=1.
Step 4. (iii)f(x)=x for x≤1, f(x)=x2 for x>1; f(1)=1.
For h→0−, 1+h≤1 so the x-branch applies: f(1+h)=1+h. Quotient =h(1+h)−1=hh=1. So f′(1−)=1.
For h→0+, 1+h>1 so the x2-branch applies: f(1+h)=(1+h)2=1+2h+h2. Quotient =h1+2h+h2−1=h2h+h2=2+h→2. So f′(1+)=2.
Since 1=2, f is not differentiable at x=1 (even though f is continuous there).
✓Final answer
None of the three functions is differentiable at x=1: (i) LHD=−1=RHD=1;
(ii) LHD→−∞ and RHD is undefined;
(iii) LHD=1=RHD=2.
Trying to compute f′(1+) for 1−x2 even though the function isn't defined for x>1
Not simplifying −h(2+h) correctly for h<0 (sign errors under the square root)
Using the wrong piecewise branch when substituting f(1+h) for h<0 vs h>0