Sol. We have,

f(x) =

∴ | f(x)| =

=

Since x ∈ [–3, 3]. Therefore, |x| ∈ [0, 3]

Thus,

f(|x|) = |x| –2 for x ∈ [–3, 3]

=

∴ g(x) = f(|x|) + |f(x)|

=

Clearly, g(x), being a polynomial or constant function, is

differentiable on (–3, 0) ∪ (0, 2) ∪ (2, 3). So, the only possible points where g(x) may not be differentiable are x = 0 and x = 2.

Differentiability at x = 0, We have, (LHD at x= 0)

=

= = –1

and,

(RHD at x= 0)

=

= = 0

∴ (LHD at x = 0) ≠ (RHD at x = 0)

So, g(x) is not differentiable at x = 0.

Differentiability at x = 2

(LHD at x = 2)

=

=

= 0

and,

(RHD at x = 2)

=

=

= 2

= 2

∴ (L.H.D. at x = 2) ≠ (RHD at x = 2)

Hence, g(x) is not differentiable at x = 2.