Answer to Question #346691 in Real Analysis for Nikhil

Question

True or false with full explanation

i. Every continuous function is differentiable.

ii. Every integrable function is monotonic

Answer 1

ANSWER Both statements are false.

EXPLANATION

Let

$f(x)=\begin{cases} & \text -2 x+1,\, \, \, { if } \, -1\leq x<0 \\ & \text0.4 x+1,\, \, { if } \, \,\, \, 0\leq x\leq 1 \end{cases}$

Since the function coincides with the polynomials on the segments $[-1,0)$ and $(0,1]$ then the function is continuous on these segments.

$\lim_{x\rightarrow 0^{-}}f(x)= \lim_{x\rightarrow 0^{-}}(-2x+1)=1=f(0) =\\=\lim_{x\rightarrow 0^{+}}f(x)= \lim_{x\rightarrow 0^{+}}(0.4x+1 )=1 .$

Thus, $f$ is continuous at the point $x=0$ and on the segment $[-1,1].$

(i)

The left-hand derivative at $x=0$ is

$f _{-}^{'}(0)=\lim_{x\rightarrow 0^{-}}\frac{f(x)-f(1) }{x}=\lim_{x\rightarrow 0^{-}}\frac{-2x+1-1 }{x}=-2$ ,

the right-hand derivative at $x=0$ is

$f _{+}^{'}(0)=\lim_{x\rightarrow 0^{+}}\frac{f(x)-f(1) }{x}=\lim_{x\rightarrow 0^{+}}\frac{0.4x+1-1 }{x}=0.4$

Since $f_{-}^{'}(0)\neq f_{+}^{'}(0)$ , then function is not differentiable at the point $x=0$

(ii) Because function $f$ is continuous on the segment $[-1,1]$ , then the function is integrable

on $[-1,1].$ But this function is not monotonic on the segment $[-1,1]$ , since in the interval $(-1,0)$ the function decreases , and in the interval $(0,1)$ the function increases .