Answer to Question #253735 in Calculus for Question

Question

Let f(x) = |2x à ƒ ƒ ¢ ˆ ’ 10| + 2 .
a) Show that f is continuous at x = 5. [8]
b) Show that f is not differentiable at x = 5. [6]

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 $x=5.$