Question #602c5

1 Answer
Oct 14, 2016

I will assume there is no error in the question.

Explanation:

f(x) = root(3)(3x-6+5) = root(3)(3x-1) = (3x-1)^(1/3)f(x)=33x6+5=33x1=(3x1)13

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

= 1/(3(3x-1)^2/3) * 3

= 1/(root(3)(3x-1))^2

f'(x) exists for all x except solutions to (root(3)(3x-1))^2 =0

f'(x) exists for all x except 1/3

If the argument of the 3rd root should be 3x^2-6x+5, then the function is differentiable at every real number.

Because 3x^2-6x+5 = 0 has no real number solution.