Find all values of x for which the function is differentiable. y=cuberoot(3x-6)+5

Am I supposed to set the derivative equal to 0??
I have the derivative.. y'=(1/3)(3x-6)^(-2/3).. now what do I do with this?

Question

Find all values of x for which the function is differentiable. y=cuberoot(3x-6)+5

Am I supposed to set the derivative equal to 0??
I have the derivative.. y'=(1/3)(3x-6)^(-2/3).. now what do I do with this?

myininaya · Answer

find the domain of y'

anonymous · Answer

The domain is all real numbers, right?

myininaya · Answer

$$y=\sqrt[3]{3x-6}+5=> y'=\frac{1}{3}(3x-6)^{\frac{1}{3}-\frac{3}{3}}+0$$
$$y'=\frac{1}{3}(3x-6)^\frac{-2}{3}=\frac{1}{3(3x-6)^\frac{2}{3}}$$

3x-6 can't be zero, right?

anonymous · Answer

Ohh yes, I forgot about the negative exponent. So I find when 3(3x-6)^(2/3) = 0?

myininaya · Answer

find the domain of y'

myininaya · Answer

all real numbers except when 3x-6=0

which is when x=2

myininaya · Answer

all real numbers except x=2