Question

Let f(x) = |2x − 10| + 2 .
a) Show that f is continuous at x = 5
b) Show that f is not differentiable at x = 5

Answer 1

a)

Let $\epsilon>0$ be given. We must find a $\delta=\delta(\epsilon) >0$ such that if$|x-5|<\delta$ then, $|f(x)-f(5)|<\epsilon.$

Now,

$|f(x)-f(5)|=||2x-10|+2-2|=||2x-10||=2|x-5|<2\delta =\epsilon \, \text{if} \,\delta=\epsilon/2\\ \implies |f(x)-f(5)|<\epsilon$

Hence, given $\epsilon>0,$ choose $\delta=\epsilon/2.$ Then, $|f(x)-f(5)|<\epsilon$ whenever $|x-5|<\delta,$ so that f is continuous at $x=5.$

b)

To show that f is not differentiable at x=5, we must show that the left hand derivative is not equal to the right hand derivative at 5. That is, $f'(5^-) \neq f'(5^+).$

$f'(5^-)=\lim_{x \to 5^-}\frac{f(x)-f(5)}{x-5}=\lim_{h \to 0}\frac{f(5-h)-2}{5-h-5}=\lim_{h \to 0}\frac{2+2h-2}{-h}=\lim_{h \to 0}\frac{2h}{-h}=-2$

$f'(5^+)=\lim_{x \to 5^+}\frac{f(x)-f(5)}{x-5}=\lim_{h \to 0}\frac{f(5+h)-2}{5+h-5}=\lim_{h \to 0}\frac{2+2h-2}{h}=\lim_{h \to 0}\frac{2h}{h}=2$

$-2\neq2 \implies f'(5^-)\neq f(5^+)$

Hence, f is not differentiable at x=5.