Dtyjlui

Is the following statement is true/false ?

Given $f : mathbb{R} rightarrow mathbb{R}$ define by
$f(x) = |x|^frac{3}{2}$

then $f(x)$ is not differentiable not at $0$?

I thinks this statetment is true because $f(x) =|x|$ is not differentible at $0$

Consider the basic definition of a function being differentiable at a given point. In this case, notice that

$$lim_{h to 0^{+}} cfrac{f(h) - f(0)}{h - 0} = lim_{h to 0^{+}} cfrac{lvert h rvert^{frac{3}{2}}}{h} = lim_{h to 0^{+}} cfrac{h^{frac{3}{2}}}{h} = lim_{h to 0^{+}} h^frac{1}{2} = 0 tag{1}label{eq1}$$

$$lim_{h to 0^{-}} cfrac{f(h) - f(0)}{h - 0} = lim_{h to 0^{-}} cfrac{lvert h rvert^{frac{3}{2}}}{h} = lim_{h to 0^{-}} cfrac{left(-hright)^{frac{3}{2}}}{h} = lim_{h to 0^{-}} -left(-hright)^frac{1}{2} = 0 tag{2}label{eq2}$$

Putting eqref{eq1} and eqref{eq2} together shows that

$$f^{'}left(0right) = lim_{h to 0} cfrac{f(h) - f(0)}{h - 0} = 0 tag{3}label{eq3}$$

Thus, $f$ is differentiable at $0$.

In general, consider

$$fleft(xright) = lvert x rvert^c, c in R tag{4}label{eq4}$$

Adjusting eqref{eq1} and eqref{eq2} gives left & right side limits as

$$lim_{h to 0^{+}} h^{c - 1} text{ and } lim_{h to 0^{-}} - left(-hright)^{c - 1} tag{5}label{eq5}$$

If $c = 1$, then the limit values become $pm 1$, so $f$ is not differentiable at $0$, as noted in the question. For $c < 1$, the limits become $pm infty$, so once again the function is not differentiable at $0$. However, for any $c > 1$, the function is differentiable at $0$, as you can verify from

$$lim_{h to 0^{+}} h^{c - 1} = lim_{h to 0^{-}} - left(-hright)^{c - 1} = 0 tag{6}label{eq6}$$

This, of course, includes not only the $c = frac{3}{2}$ asked for specifically in the question, but also for $c = 2$ mentioned by Randall in the question comments (note this makes the function the equivalent of $fleft(xright) = x^2$).

In most cases, including this one, a somewhat easier approach would be to determine the derivative of the function for values of $x < 0$ and the limit as $x to 0^{-}$, then the derivative of the function for $x > 0$ and the limit as $x to 0^{+}$. If both limits exist and are equal to each other, then the function is differentiable at $0$. For this specific example, for $x < 0$, $f_{1}left(xright) = left(-xright)^{frac{3}{2}}$ so $f_{1}^{'}left(xright) = -frac{3}{2}left(-xright)^{frac{1}{2}}$, giving that $lim_{x to 0^{-}} f_{1}^{'}left(xright) = 0$. For $x > 0$, $f_{2}left(xright) = x^{frac{3}{2}}$ so $f_{2}^{'}left(xright) = frac{3}{2}x^{frac{1}{2}}$, giving that $lim_{x to 0^{+}} f_{2}^{'}left(xright) = 0$. As both limits are $0$, $fleft(xright)$ is differentiable at $x = 0$.