Question

Let f(x) = |2x − 10| + 2 . a) Show that f is continuous at x = 5. [8] b) Show that f is not differentiable at x = 5

Answer 1

$$f(x) = |2x-10| + 2$$

a)

$$\lim\limits_{x\to5^-}f(x)=\lim\limits_{x\to5^-}(|2x-10| + 2 )$$

$$=\lim\limits_{x\to5^-}(10-2x + 2 )=12-2(5)=2$$

$$\lim\limits_{x\to5^+}f(x)=\lim\limits_{x\to5^+}(|2x-10| + 2 )$$

$$=\lim\limits_{x\to5^+}(2x-10 + 2 )=2(5)-8=2$$

Then

$$\lim\limits_{x\to5^-}f(x)=2=\lim\limits_{x\to5^+}f(x)=>\lim\limits_{x\to5}f(x)=2$$

$$f(5)=|2(5)-10|+2=0+2=2$$

We have

$$\lim\limits_{x\to5}f(x)=2=f(5)$$

Therefore the function $f(x)$ is continuous at $x=5.$

b)

$$f(5)=|2(5)-10|+2=0+2=2$$ $$\lim\limits_{h\to0^-}\dfrac{f(5+h)-f(5)}{h}=\lim\limits_{h\to0^-}\dfrac{|2(5+h)-10|+2-2}{h}$$

$$=\lim\limits_{h\to0^-}\dfrac{2|h|}{h}=\lim\limits_{h\to0^-}\dfrac{-2h}{h}=-2$$

$$\lim\limits_{h\to0^+}\dfrac{f(5+h)-f(5)}{h}=\lim\limits_{h\to0^+}\dfrac{|2(5+h)-10|+2-2}{h}$$

$$=\lim\limits_{h\to0^+}\dfrac{2|h|}{h}=\lim\limits_{h\to0^+}\dfrac{2h}{h}=2$$

Since

$$\lim\limits_{h\to0^-}\dfrac{f(5+h)-f(5)}{h}=-2$$

$$\not=2=\lim\limits_{h\to0^+}\dfrac{f(5+h)-f(5)}{h},$$

then $\lim\limits_{h\to0}\dfrac{f(5+h)-f(5)}{h}$ does not exist.

Therefore the function $f(x)$ is not differentiable at $x=5.$