The set of points where the function f(x) = x |x| is differentiab… - Vidyadip

Question

The set of points where the function f(x) = x |x| is differentiable is:

$(-\infty,\infty)-\infty,\infty$
$(-\infty,0)\cup(0,\infty)-\infty,0\cup0,\infty$
$(0,\infty)0,\infty$
$[0,\infty]0,\infty$

✓

Answer

$(-\infty,\infty)$

Solution:

We have,

$\text{f(x)}=\text{x}|\text{x}|$

$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}^2, & \text{x}<0\\0 ,& \text{x}= 0\\\text{x}^2,&\text{x}>0\end{cases}$

When, x < 0, we have

f(x) = -x² which being a polynomial function is continuous and differentable in $(-\infty,0)$

When, x > 0, we have

f(x) = -x² which being a polynomial function is continuous and differentable in $(0,\infty,)$

Thus possible point of non-differentiability of f(x) is x = 0

Now, LHL (at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{x}^2-0}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-(-\text{h})^2}{-\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\text{h}$

$=0$

And RHL (at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{x}^2-0}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\text{h}$

$=0$

$\therefore$ LHL (at x = 0) = RHL (at x = 0)

So, f(x) is also differentiable at x = 0

i.e. f(x) is differentiable in $(-\infty,\infty).$

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