Question #242618

discuss the continuity/uniform continuity/Lipchitz continuity and differentiability of the functions |x| on [−1,1].


1
Expert's answer
2021-09-27T14:31:49-0400

ANSWER.Function f(x)=xf(x)=|x| is Lipschitz continuous, therefore, uniformly continuous, hence continuous on [-1,1]. The function is not differentiable at all points of the [-1,1] (At the point x=0 the function is not differentiable).

EXPLANATION.

Function f(x)=xf(x)=|x| is Lipschitz continuous, because for all x1,x2[1,1]: f(x1 )f (x2)=x1 x2  x1x2{ x }_{ 1 },{ x }_{ 2 }\in \left[ -1,1 \right] :\ \left| f({ x }_{ 1\ })-f\ ({ x }_{ 2 }) \right| =\left| \left| { x }_{ 1\ } \right| -\left| { x }_{ 2\ } \right| \right| \ \le \left| { x }_{ 1 }-{ x }_{ 2 } \right| (1).

Since for any ε>0ε>0 and δ=εδ=ε for any x1,x2[1,1]{ x }_{ 1 },{ x }_{ 2 }\in \left[ -1,1 \right] such that x1x2<δ\left| { x }_{ 1 }-{ x }_{ 2 } \right| <\delta \quadfrom (1) it follows f(x1 )f (x2)=x1 x2 <ε\left| f({ x }_{ 1\ })-f\ ({ x }_{ 2 }) \right| =\left| \left| { x }_{ 1\ } \right| -\left| { x }_{ 2\ } \right| \right| < ε , then ff is uniformly continuous.

ff is differentiable on [1,0)[-1,0) and (0,1](0,1] , because

f(x)={x,1x<00,x=0x,0<x1,f(x)={1,1x<01 ,0<x1f(x)=\begin{cases} -x,-1\le x<0 \\ 0,\quad x=0 \\ x,\quad 0<x\le 1 \end{cases}, f'(x)=\begin{cases} -1,-1\le x<0 \\ 1\ ,\quad 0<x\le 1 \end{cases} .

Since f(0)=limx0f(x)f(0)x=limx0x0x=1,f+(0)=limx0+f(x)f(0)x=limx0+ x0x= 1{ f }_{ - }^{ ' }(0)=\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { f(x)-f(0) }{ x } } =\lim _{ x\rightarrow { 0 }^{ - } }{ \frac { -x-0 }{ x } =-1,\quad { f }_{ + }^{ ' }(0)= } \lim _{ x\rightarrow { 0 }^{ + } }{ \frac { f(x)-f(0) }{ x } } =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { \ x-0 }{ x } =\ 1 }

then the left derivative is not equal right derivative. So , the function is not differentiable at the point x=0.x=0.


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