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

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Homework Help Overview

The discussion revolves around the differentiability of the function |x|^3 at the point x=0. Participants are examining the definition of the derivative and the behavior of the function near this point.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the limits defining the derivative from both sides of x=0, questioning the implications of differing left and right limits. Some participants attempt to clarify their calculations regarding these limits.

Discussion Status

There is an ongoing examination of the derivative's existence at x=0, with some participants asserting differentiability based on their calculations, while others express uncertainty and seek further clarification. A mix of interpretations regarding the limits and their implications is present.

Contextual Notes

Some participants mention potential confusion in their calculations and the need to clarify the definitions and properties of the function involved.

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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<t<b, \ t\neqx), \\

and \ define \\

f^{'}(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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