Describe the x-values at which f is differentiable. (Enter your answer using interval notation.) f(x) = (x + 3)^2/3

Question

The ^ is an exponent the exponent i 2/3

Answer 1

Find f'(x)

f(x) = (x + 3)^2/3

f'(x) = (2/3)(x+3)^-1/3

f'(x) = (2/3)[1/(x+3)^1/3]

f'(x) = 2/[3(x+3)^1/3]

Since 3(x+3)^1/3≠0, then x≠-3, making the x-values at which f is differentiable (-∞,-3)υ(-3,∞). If you graph the derivative, you can also see this for yourself because of the vertical asymptote at x=-3.

Hope this helped!