Question #602c5

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Answer 1 · 2016-10-14T18:01:49

I will assume there is no error in the question.

$f (x) = \sqrt[3]{3 x - 6 + 5} = \sqrt[3]{3 x - 1} = {(3 x - 1)}^{\frac{1}{3}}$ $f(x) = root(3)(3x-6+5) = root(3)(3x-1) = (3x-1)^(1/3)$

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

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

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

$f'(x)$ 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.