Answer
The graph of the given function has a vertical tangent at the origin.
Work Step by Step
$y=f(x)=-\sqrt{|x|}$ for $x\le0$ and $y=f(x)=\sqrt x$ for $x\gt0$
a) The graph is enclosed below. From the introduction, we can guess that the graph appears to have vertical tangent at $x=0$. We will test this in part b).
b) The graph of $f(x)$ has a vertical tangent at $x$ when
1) $f$ is continuous at $x$
2) $$f′(x)=\lim_{h\to0}\frac{f(x+h)−f(x)}{h}=\pm\infty$$
Considering $f'(0)$, we have:
$$f′(0)=\lim_{h\to0}\frac{f(h)−f(0)}{h}$$
- For $f(0)$: $f(0)=-\sqrt{|0|}=0$
- For $f(h)$: since there are 2 functions for $f(h)$, we need to consider 2 cases:
$$\lim_{h\to0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to0^-}\frac{-\sqrt{|h|}-0}{h}=\lim_{h\to0^-}\frac{-\sqrt{|h|}}{h}$$
Since $h\lt0$ here, $|h|=-h$: $$\lim_{h\to0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to0^-}\frac{-\sqrt{-h}}{-(-h)}=\lim_{h\to0^-}\frac{\sqrt{-h}}{-h}=\lim_{h\to0^-}\frac{1}{\sqrt{-h}}=\infty$$
$$\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to0^+}\frac{\sqrt h-0}{h}=\lim_{h\to0^+}\frac{\sqrt h}{h}=\lim_{h\to0^+}\frac{1}{\sqrt h}=\infty$$
Therefore, $$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\infty$$
This means $f$ has a vertical tangent at the origin as predicted.