Answer
The graph of $y=\sqrt{|4-x|}$ does not have any vertical tangents anywhere.
Work Step by Step
$$y=f(x)=\sqrt{|4-x|}$$
a) The graph is enclosed below. It does not seem to have any vertical tangents, even at $x=4$, because as we see from the introduction that the graph of $y=x^{2/3}$ does not have a vertical tangent at the origin.
b) The graph of $f(x)$ has a vertical tangent at $x$ when
1) $f$ is continuous at $x$
2) $$f′(x)=\lim_{h\to0}\frac{f(x+h)−f(x)}{h}=\pm\infty$$
Here, we can see that:
$$f′(4)=\lim_{h\to0}\frac{f(h+4)−f(4)}{h}=\lim_{h\to0}\frac{\sqrt{|4-(h+4)|}-\sqrt{|4-4|}}{h}$$
$$f'(4)=\lim_{h\to0}\frac{\sqrt{|-h|}-\sqrt0}{h}=\lim_{h\to0}\frac{\sqrt{|-h|}}{h}$$
- As $h\to0^+$, $-h\lt0$, so $|-h|=h$. That means $$\lim_{h\to0^+}\frac{\sqrt{|-h|}}{h}=\lim_{h\to0^+}\frac{\sqrt h}{h}=\lim_{h\to0^+}\frac{1}{\sqrt h}=\infty$$
- As $h\to0^-$, $-h\gt0$, so $|-h|=-h$. That means $$\lim_{h\to0^-}\frac{\sqrt{|-h|}}{h}=\lim_{h\to0^-}\frac{\sqrt {-h}}{h}=\lim_{h\to0^-}-\frac{\sqrt{-h}}{-h}=-\lim_{h\to0^-}\frac{1}{\sqrt{-h}}=-\infty$$
This means $\lim_{h\to0}\frac{\sqrt{|-h|}}{h}$ does not exist. Therefore, as predicted, $f$ does not have a vertical tangent at $x=4$.
