Question

Question #602c5

Answer 1

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) = 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.