Is |x|^3 Differentiable at x=0?

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SUMMARY

The function |x|^3 is differentiable at x=0, with the derivative at that point equal to 0. Both the left-hand limit and right-hand limit of the derivative as t approaches 0 yield a value of 0, confirming differentiability. The calculations show that for t>0, the limit of (|t^3|)/(t) approaches 0, and for t<0, the limit of (-t^3)/(t) also approaches 0. Therefore, |x|^3 is differentiable everywhere, including at x=0.

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[SOLVED] Is |x|^3 differentiable?

Homework Statement



Is |x|^3 differentiable?

Homework Equations



Def: \ Let \ f \ be \ defined \ (and \ real-valued) \ on [a,b]. \ \ For \ any \ x \in [a,b], \ form \ the \ quotient

\phi(t)=\frac{f(t)-f(x)}{t-x} \ \ \ \ (a&lt;t&lt;b, \ t\neqx), \\

and \ define \\

f^{&#039;}(x)=\lim_{\substack{t\rightarrow x}} \phi(t)

The Attempt at a Solution



Well, using the definition, I calculated that the right-hand limit and left hand limit are different. But I'm not sure if that means anything or what I can conclude here. Nor am I sure how I should define the left and right limits here.

Any help would be great, thanks!
 
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If the left hand limit and the right hand limit are different, then the limit that defines the derivative at that point does not exists: the function is not differentiable.

However, I'm not sure it's not differentiable in this case... if there were a problem it would be at x = 0 and when I sketch a picture I don't immediately see a problem arising. Can you show us the calculation for the limit?
 
The left hand derivative and the right hand derivative at x=0 are both zero. It's differentiable.
 
CompuChip said:
If the left hand limit and the right hand limit are different, then the limit that defines the derivative at that point does not exists: the function is not differentiable.
How did you get that the right and left hand derivatives are different? If t> 0, |f(t)|= |t^3|= t^3 then at x= 0,
\frac{f(t)-f(x)}{t- x}= \frac{|t^3|}{t}= \frac{t^3}{t}= t^2
and the limit of that, as t goes to 0, is 0.

If t< 0, |f(t)|= |t^3|= -t^3. At x= 0,
\frac{f(t)-f(x)}{t-x}= \frac{|t^3|}{t}= \frac{-t^3}{t}= -t^2
but the limit, as t goes to 0, is still 0. The function is differentiable at 0 and the derivative there is 0.

Obviously, if x> 0, f(x)=|x^3|= x^3 which is differentiable and if x< 0, f(x)= |x^3|= -x^3 which is differentiable.
 
Ah, I see what I did now. Totally forgot to cube my expression in my calculations, hahaha. Alrighty, thanks Ivy.
 
|x|^3 the same as |x^3| [\tex] ?
 
They are equal, not same. They are different functions.
 

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