Explanation

(a)
We have, $f(x) = \left\{ \begin{array}{r}
x^{2},\ \ \ x \geq 0 \\
- x^{2},\ \ \ \ x < 0
\end{array} \right.\$
Clearly, $f(x)$ is differentiable for all $x > 0$ and for all$x < 0$. So, we check the differentiable at $x = 0$
Now, (RHD at $x = 0$)
$\left( \frac{d}{dx}(x)^{2} \right)_{x = 0} = (2x)_{x = 0} = 0$
And (LHD at $= 0$)
$\left( \frac{d}{dx}( - x)^{2} \right)_{x = 0} = ( - 2x)_{x = 0} = 0$
$\therefore\ \ \$(LHD at $x = 0$)=(RHD at $x = 0$)
So, $f(x)$ is differentiable for all $x\ ie$, the set of all points where $f(x)$ is differentiable is $( - \infty,\ \infty)$
Alternate
It is clear from the graph $f(x)$ is differentiable everywhere.
[image:5_image6.emf]