Determine whether f is differentiable at the given point.

if f '(a) exists, give its value. hint [see example 3.] (if an answer does not exist, enter dne.)

f(x) = x^1/5 + 7

a= 32

a=0

I hope this helps you f'(x)=1/5. (x)^1/5-1 f'(x)=1/5.(x)^1-5/5 f'(x)=1/5.(x)-4/5 a=32=2^5 f'(32)=f'(2^5)=1/5. (2^5)^-4/5 f'(2^5)=1/5.2^5.-4/5 f'(2^5)=1/5.2^-4 f'(2^5)=1/5.2^4 f'(2^5)=1/80 a=0 f'(0)= 1/5. (0)^-4/5 f'(0)= 1/5.0 f'(0)=0

D. i think it is the savings

The answer is a hope i