Answer
given,
[tex]f(x) = 2 x^{1/5}[/tex]
interval = [-32, 32]
differentiating the given equation
[tex]f'(x) = \dfrac{d}{dx}(2 x^{1/5})[/tex]
[tex]f'(x) = \dfrac{2}{5x^{4/5}}[/tex]
hence, the above solution is not defined at x = 0
and x = 0 lie in the given interval i.e. [-32, 32]
so, at x = 0 the function is not differentiable.
Hence, mean value theorem does not apply to the given function.
