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Differentiability at a point via one-sided derivative behavior

Consider the piecewise function f(x)={−xx<0x1/3x≥0f(x) = \begin{cases} -x & x < 0 \\ x^{1/3} & x \geq 0 \end{cases} We want to determine whether ff is differentiable at x=0x = 0.

Using the limit definition of the derivative, examine the behavior of f(0+h)−f(0)h\frac{f(0+h) - f(0)}{h} as h→0−h \to 0^- and as h→0+h \to 0^+. Based on this, is ff differentiable at x=0x = 0? Justify your conclusion by describing what feature (corner, cusp, vertical tangent, or discontinuity) occurs at x=0x=0.

Differentiability at a point via one-sided derivative behavior graph
Hint

Compute the difference quotient f(h)−f(0)h\frac{f(h)-f(0)}{h} separately for h<0h<0 using f(h)=−hf(h)=-h, and for h>0h>0 using f(h)=h1/3f(h)=h^{1/3}, then take each one-sided limit as h→0h\to 0.

Step-by-step solution

For h<0h < 0, f(0)=0f(0)=0 and f(h)=−hf(h) = -h, so the difference quotient is −h−0h=−1\frac{-h - 0}{h} = -1. Thus lim⁡h→0−f(h)−f(0)h=−1\lim_{h\to 0^-}\frac{f(h)-f(0)}{h} = -1.

For h>0h > 0, f(h)=h1/3f(h) = h^{1/3}, so the difference quotient is h1/3h=h−2/3=1h2/3\frac{h^{1/3}}{h} = h^{-2/3} = \frac{1}{h^{2/3}}. As h→0+h \to 0^+, h2/3→0+h^{2/3}\to 0^+, so this quotient →+∞\to +\infty.

Since the left-hand limit is −1-1 (finite) and the right-hand limit is +∞+\infty, the limit defining f′(0)f'(0) does not exist. The left piece has slope −1-1, while the right piece x1/3x^{1/3} has a vertical tangent at x=0x=0 (where the derivative approaches +∞+\infty). This creates a cusp at x=0x=0, so ff is not differentiable at x=0x=0 even though it is continuous there.

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