Let $\displaystyle f:R-->R$ be a function defined as

$\displaystyle f(x)=x^{3/2}$,x>=0

$\displaystyle -|x|^{3/2}$, x<0.

What can be said about its differentiability? Is is differentiable at 0? If yes, how many times?

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- Feb 20th 2011, 07:34 AMSambitdifferentiability
Let $\displaystyle f:R-->R$ be a function defined as

$\displaystyle f(x)=x^{3/2}$,x>=0

$\displaystyle -|x|^{3/2}$, x<0.

What can be said about its differentiability? Is is differentiable at 0? If yes, how many times? - Feb 20th 2011, 08:09 AMFernandoRevilla
Verify $\displaystyle f'_{+}(0)=f'_{-}(0)=0$ so , :

$\displaystyle f'(x)=\begin{Bmatrix}{ 3\sqrt{x}/2}&\mbox{ if }& x>0\\0 & \mbox{if}& x=0\\\ldots & \mbox{if}& x<0\end{matrix} $

Now, you can study the existence or not of $\displaystyle f''(0)$ .

Fernando Revilla - Feb 20th 2011, 05:54 PMSambit
I am getting $\displaystyle f(x)=x^{3/2}$,x>=0

$\displaystyle -x^{3/2}$,x<0

which is differentiable.

Differentiating, $\displaystyle f'(x)=\frac{3}{2}x^{1/2}$, x>=0

$\displaystyle \frac{3}{2}(-x)^{1/2}$,x<0

This is also continuous and differentiable at 0, and so on. So I think it is correct that the function is infinitely differentiable. Am I right? - Feb 20th 2011, 11:33 PMFernandoRevilla

No, prove that:

$\displaystyle \displaystyle\lim_{h \to 0^+}{\dfrac{f'(h)-f'(0)}{h}}=\ldots=+\infty$

so, $\displaystyle f''(0)$ does not exist.

Fernando Revilla - Feb 21st 2011, 03:17 AMSambit
Okay....got it. thank you.