Question

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

Answer 1

\[\sqrt[3]{3x-6}+5\]

Answer 2

d/dx=(1/3)*(3x-6)^(-2/3)*3
=(3x-6)^(-2/3)
then it will be differentiabel over x=R-{2}

Answer 3

Huh? I get up to this part..\[(1/3)(3x-6)^{(-2/3)}\]

So how do I find the values for which x is differentiable?

Answer 4

you need to write what that thing really is

Answer 5

it is
\[\frac{1}{3\sqrt[3]{(3x-6)^2}}\]

Answer 6

and since you have a denominator it cannot be zero, in other words
\[3x-6\neq 0\]
\[x\neq2\]

Answer 7

I think I need to go back to algebra again...
What happened to the square and the cuberoot?

Answer 8

\[b^{\frac{p}{q}}=\sqrt[q]{b^p}\]

Answer 9

Oh yes I understand that part.

Answer 10

and the minus sign puts it in the denominator

Answer 11

oh maybe you mean what happened to them when i solved?

Answer 12

Ohh nevermind, I see now. ?