Differenriable of a function y=x^3/(abs(x)) at x=0

[Arun0293]

Function y=x^3/(abs(x))
I got a parabola after plotting the function
y= x^2 if x>0
y=-x^2 if x<0
Can anyone tell whether the function is differentiable at x=0? Please justify using left hand side limits and right hand side limits.

 

[stapel]

Function y=x^3/(abs(x))
I got a parabola after plotting the function
y= x^2 if x>0
y=-x^2 if x<0
Can anyone tell whether the function is differentiable at x=0? Please justify using left hand side limits and right hand side limits.

Okay. What did you find, when you took the left and right limits?

Please be complete. Thank you!

 

[HallsofIvy]

The function f(x) is differentiable at 0 if and only if \(\displaystyle \lim_{h\to 0}\frac{f(h)- f(0)}{h}\) exists. And that exists if and only if \(\displaystyle \lim_{h\to 0^-}\frac{f(h)- f(0)}{h}\) and \(\displaystyle \lim_{h\to 0^+}\frac{f(h)- f(0)}{h}\) both exist and are equal. Here, \(\displaystyle f(x)= \frac{x^3}{|x|}\) which, as you have correctly deduced, is \(\displaystyle \frac{x^3}{x}= x^2\) if x is positive and \(\displaystyle \frac{x^3}{-x}= -x^2\) if x is negative.

So you need to find \(\displaystyle \lim_{h\to 0} h^2\) and \(\displaystyle \lim_{h\to 0} -h^2\). What are those limits? Do they both exist? Are they equal?

 

[Arun0293]

limx→0 x^2 =0 and limx→0 (-x^2)=0

Okay. What did you find, when you took the left and right limits?

Please be complete. Thank you!

limx→0+ (x^2) =0 and limx→0- (-x^2)=0
Both left and right hand side limits approaches zero. And the curve is continuous .

Is it differentiable at x=0?

 

[HallsofIvy]

Read my previous response!

 

[Arun0293]

Differentable

Read my previous response!

LHS
limx→0− (f(-0.1)-f(0))/(-0.1-0) = (-(-0.1^2)-0^2)/-0.1= 0.01/0.1 = 0.1

RHS

limh→0+limh→0+

limh→0−

limx→0+ (f(0.1)-f(0))/(0.1-0) = (0.1^2-0^2)/(0.1) = 0.01/0.1 = 0.1

So L'f(0) = R'f(0) so it differentiable at x = 0 . Am i right?