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Vertical tangent at x = 0

Consider the function f(x)={xx≥0−−xx<0f(x) = \begin{cases} \sqrt{x} & x \geq 0 \\ -\sqrt{-x} & x < 0 \end{cases}

Use the limit definition f′(0)=lim⁡h→0f(h)−f(0)hf'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} to determine whether ff is differentiable at x=0x = 0. If it is, state the value of f′(0)f'(0); if it is not, explain why the limit fails to exist (as a finite number).

Vertical tangent at x = 0 graph
Hint

Since f(0)=0f(0) = 0, the difference quotient is f(h)/hf(h)/h. Compute this for h>0h > 0 and for h<0h < 0 separately, then examine what happens as h→0+h \to 0^+ and h→0−h \to 0^-. Watch what happens to the size of the quotient, not just its sign.

Step-by-step solution

For h>0h > 0, f(h)=hf(h) = \sqrt{h}, so the difference quotient is hh=1h\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}. As h→0+h \to 0^+, h→0+\sqrt{h} \to 0^+ so 1h→+∞\frac{1}{\sqrt{h}} \to +\infty.

For h<0h < 0, f(h)=−−hf(h) = -\sqrt{-h}, so the difference quotient is −−hh\frac{-\sqrt{-h}}{h}. Since h<0h < 0, we have h=−∣h∣h = -|h|, so −−hh=−∣h∣−∣h∣=∣h∣∣h∣=1∣h∣\frac{-\sqrt{-h}}{h} = \frac{-\sqrt{|h|}}{-|h|} = \frac{\sqrt{|h|}}{|h|} = \frac{1}{\sqrt{|h|}}. This also tends to +∞+\infty as h→0−h \to 0^-. Since both one-sided limits diverge to +∞+\infty, the limit defining f′(0)f'(0) does not exist as a finite number, so ff is not differentiable at x=0x=0. The graph shown in Vertical tangent at x = 0 — solution graph illustrates this vertical tangent behavior at the origin, where ff is continuous but not differentiable.

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