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Differentiability at x = 0

Consider the function f(x)=∣x∣+∣x−1∣f(x) = |x| + |x-1|, which is a sum of two absolute value expressions. Its graph has corner points where each absolute value term changes behavior, namely at x=0x=0 and x=1x=1.

Consider f(x)=∣x∣+∣x−1∣f(x) = |x| + |x-1|.

Differentiability at x = 0 graph

Using the limit definition of the derivative, f′(0)=lim⁡h→0f(0+h)−f(0)hf'(0) = \lim_{h \to 0} \dfrac{f(0+h) - f(0)}{h}, examine the one-sided limits as h→0+h \to 0^+ and h→0−h \to 0^-. Is ff differentiable at x=0x = 0? Justify your answer by stating what happens to the difference quotient from each side.

Hint

Near x=0x=0, split into cases h>0h>0 and h<0h<0 to remove the absolute value signs, since x−1x-1 stays negative for hh near 0. Compare the resulting slopes from each side.

Step-by-step solution

For h>0h>0 small, f(h)=h+(1−h)=1f(h) = h + (1-h) = 1, and f(0)=0+1=1f(0) = 0 + 1 = 1, so the difference quotient is f(h)−f(0)h=1−1h=0\frac{f(h)-f(0)}{h} = \frac{1-1}{h} = 0. Thus the right-hand limit of the difference quotient is 00.

For h<0h<0 small, f(h)=−h+(1−h)=1−2hf(h) = -h + (1-h) = 1 - 2h, so f(h)−f(0)h=(1−2h)−1h=−2hh=−2\frac{f(h)-f(0)}{h} = \frac{(1-2h) - 1}{h} = \frac{-2h}{h} = -2. Since the left-hand limit (−2-2) does not equal the right-hand limit (00), the two-sided limit does not exist, so ff is not differentiable at x=0x=0 — the graph has a corner there.

Differentiability at x = 1

Using the same function f(x)=∣x∣+∣x−1∣f(x) = |x| + |x-1|, use the limit definition of the derivative to examine f′(1)=lim⁡h→0f(1+h)−f(1)hf'(1) = \lim_{h \to 0} \dfrac{f(1+h) - f(1)}{h} from both sides. Is ff differentiable at x=1x = 1? Justify your answer by comparing the one-sided slopes.

Hint

Near x=1x=1, the term ∣x∣|x| stays positive (equal to xx), so only ∣x−1∣|x-1| changes form depending on the sign of hh. Compute the difference quotient separately for h>0h>0 and h<0h<0.

Step-by-step solution

For h>0h>0 small, f(1+h)=(1+h)+h=1+2hf(1+h) = (1+h) + h = 1+2h, and f(1)=1+0=1f(1) = 1+0=1, so f(1+h)−f(1)h=2hh=2\frac{f(1+h)-f(1)}{h} = \frac{2h}{h} = 2. This gives a right-hand limit of 22.

For h<0h<0 small, ∣1+h−1∣=∣h∣=−h|1+h-1| = |h| = -h (since h<0h<0), so f(1+h)=(1+h)+(−h)=1f(1+h) = (1+h) + (-h) = 1. Thus f(1+h)−f(1)h=1−1h=0\frac{f(1+h)-f(1)}{h} = \frac{1-1}{h} = 0. Since the left-hand limit (00) and right-hand limit (22) disagree, the derivative does not exist at x=1x=1, so ff has a corner there as well.

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