Question #282470

following statement is true and which are false? Give reasons for your answers, in the form of a short proof or a counter example.





Every continuous function is differentiable

1
Expert's answer
2021-12-27T03:27:34-0500

False.

For example, the function f(x)=xf(x)=|x| is continuous at x=0x=0 :


limx0f(x)=0=limx0+f(x)=>limx0f(x)=0\lim\limits_{x\to 0^-}f(x)=0=\lim\limits_{x\to 0^+}f(x)=>\lim\limits_{x\to 0}f(x)=0

f(0)=0=0=limx0f(x)f(0)=|0|=0=\lim\limits_{x\to 0}f(x)

On the other hand,


limh0f(0+h)f(0)h=limh0h0h=1\lim\limits_{h\to 0^-}\dfrac{f(0+h)-f(0)}{h}=\lim\limits_{h\to 0^-}\dfrac{|h|-0}{h}=-1

limh0+f(0+h)f(0)h=limh0+h0h=1\lim\limits_{h\to 0^+}\dfrac{f(0+h)-f(0)}{h}=\lim\limits_{h\to 0^+}\dfrac{|h|-0}{h}=1

limh0f(0+h)f(0)h=1\lim\limits_{h\to 0^-}\dfrac{f(0+h)-f(0)}{h}=-1

1=limh0+f(0+h)f(0)h\not=1=\lim\limits_{h\to 0^+}\dfrac{f(0+h)-f(0)}{h}

Then limh0f(0+h)f(0)h\lim\limits_{h\to 0}\dfrac{f(0+h)-f(0)}{h} does not exist.

Therefore the function f(x)=xf(x)=|x| is not differentiable at x=0.x=0.


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