Showing That a Function Is Not Differentiable at a Point Show that [latex]f(x)=\left\{\begin{array}{ll}4 & \text { if } x<2 \\2 x & \text { if } x \geq 2\end{array}\right.[/latex] is not differentiable at x = 2.

The graph (see Figure 2.18) indicates a sharp corner at x = 2, so you might expect that the derivative does not exist. To verify this, we investigate the derivative by evaluating one-sided limits. For h > 0, note that (2 + h) > 2 and so, f(2 + h) = 2(2 + h). This gives us [latex]\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0^{+}} \frac{2(2+h)-4}{h}[/latex] [latex]=\lim _{h \rightarrow 0^{+}} \frac{4+2 h-4}{h}[/latex] Multiply out and cancel. [latex]=\lim _{h \rightarrow 0^{+}} \frac{2 h}{h}=2[/latex] Cancel common h’s. Likewise, if h < 0, (2 + h) < 2 and so, f(2 + h) = 4. Thus, we have [latex]\lim _{h \rightarrow 0^{-}} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0^{-}} \frac{4-4}{h}=0[/latex] Since the one-sided limits do not agree (0 = 2), f(2) does not exist (i.e., f is not differentiable at x = 2).

Question 2.2.7: Showing That a Function Is Not Differentiable at a Point Sh...

Calculus: Early Transcendental Functions

Showing That a Function Is Not Differentiable at a Point

Show that $f(x)=\left\{\begin{array}{ll}4 & \text { if } x<2 \\2 x & \text { if } x \geq 2\end{array}\right.$ is not differentiable at x = 2.

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The graph (see Figure 2.18) indicates a sharp corner at x = 2, so you might expect that the derivative does not exist. To verify this, we investigate the
derivative by evaluating one-sided limits. For h > 0, note that (2 + h) > 2 and so, f(2 + h) = 2(2 + h). This gives us

$\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0^{+}} \frac{2(2+h)-4}{h}$

$=\lim _{h \rightarrow 0^{+}} \frac{4+2 h-4}{h}$ Multiply out and cancel.

$=\lim _{h \rightarrow 0^{+}} \frac{2 h}{h}=2$ Cancel common h’s.

Likewise, if h < 0, (2 + h) < 2 and so, f(2 + h) = 4. Thus, we have