Answer
given,
![f(x) = 2 x^(1/5)](https://img.qammunity.org/2021/formulas/mathematics/college/v2suq9l8io2r3rlww0ek5grbfjn4yox84k.png)
interval = [-32, 32]
differentiating the given equation
![f'(x) = (d)/(dx)(2 x^(1/5))](https://img.qammunity.org/2021/formulas/mathematics/college/dttu8oz9bjqobh7plylhr2vyoqcfo4xijy.png)
![f'(x) = (2)/(5x^(4/5))](https://img.qammunity.org/2021/formulas/mathematics/college/uysngzx0lerray8n1aupwnfp64utnyldgu.png)
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.