Answer to Question #150315 in Calculus for stefanus weyulu

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) A function $f(x)$ is said to be continuous at a point $x=a$ of its domain,iff $\displaystyle {\lim_{x\to a}}f(x)=f(a)$

$\implies \displaystyle {\lim_{x\to a^-}}f(x)=\displaystyle {\lim_{x\to a^+}}f(x)=f(a)$

We know that, $\displaystyle {\lim_{x\to a^-}}f(x)=\displaystyle {\lim_{h\to 0}}f(a-h)$

and $\displaystyle {\lim_{x\to a^+}}f(x)=\displaystyle {\lim_{h\to 0}}f(a+h)$

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

It can be written as,$f(x)=(2x-10)+2; when$ $x>5$

$=-(2x-10)+2; when$ $x<5$

$=2;when$ $x=5$

According to the problem,

$\displaystyle {\lim_{x\to 5^-}}f(x)=\displaystyle {\lim_{h\to 0}}f(5-h)$

$=\displaystyle {\lim_{h\to 0}}[-[2(5-h)-10]]+2$

$=\displaystyle {\lim_{h\to 0}}(2h+2)$

$=2$

$\displaystyle {\lim_{x \to 5^+}}f(x)=\displaystyle {\lim_{h \to 0}}f(5+h)$

$=\displaystyle {\lim_{h\to 0}} [2(5+h)-10]+2$

$=\displaystyle {\lim_{h\to 0}}(2h+2)$

$=2$

And $f(5)=2$

$\therefore f(x)$ is continuous at $x=5$ .

b) A function f(x) is said to be differentiable at $x=a,$ iff $\displaystyle {\lim_{x\to a}}[f(x)-f(a)]/(x-a)$ exists finite.

$\implies$ $\displaystyle {\lim_{x\to a^-}}[f(x)-f(a)]/(x-a)=\displaystyle {\lim_{x\to a^+}}[f(x)-f(a)]/(x-a)$

$\implies \displaystyle {\lim_{h\to 0}}[f(a-h)-f(a)]/(-h)= \displaystyle {\lim_{h\to 0}}[f(a+h)-f(a)]/(h)$

Now,

$\displaystyle {\lim_{x\to 5^-}}[f(x)-f(5)]/(x-5)$

$=\displaystyle {\lim_{h\to 0}}[f(5-h)-f(5)]/(-h)$

$=\displaystyle {\lim_{h\to 0}}[(2h+2)-2]/(-h)$

$=\displaystyle {\lim_{h\to 0}}(-2)=-2$

$\displaystyle {\lim_{x\to 5^+}}[f(x)-f(5)]/(x-5)$

$=\displaystyle {\lim_{h\to 0}}[f(5+h)-f(5)]/(h)$

$=\displaystyle {\lim_{h\to 0}}[(2h+2)-2]/(h)$

$=\displaystyle {\lim_{h\to 0}}(2)=2$

$\therefore$ $\displaystyle {\lim_{x\to 5^-}}[f(x)-f(5)]/(x-5)$ $\neq$ $\displaystyle {\lim_{x\to 5^+}}[f(x)-f(5)]/(x-5)$

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