$C(x) = x^{1/5}(x + 6)$ Find Where the Function Increases and Decreases I believe this has something to do with my Algebra. I am stuck as to how to evaluate this local minimum and maximum. Also, my sign chart. Do I have too many intervals? If so why. 
 A: I think $x=0$ is not a critical point. To see this, look at the derivative $f'(x)$ you wrote in the first line: $$f'(x) = \frac{6}{5}x^{1/5} + \frac{6}{5}x^{-4/5}.$$
This function is actually not defined in $x=0$. Still one could have that $\lim_{x\to 0} f'(x) = 0$, but this is not the case. Indeed, note that $x^{-4/5} \to \infty$ as $x\to 0$. 
A: The derivative can be easily computed with the (formal) logarithmic method:
$$
\log f(x)=\frac{1}{5}\log x+\log(x+6)
$$
so
$$
\frac{f'(x)}{f(x)}=\frac{1}{5x}+\frac{1}{x+6}=\frac{6(x+1)}{5x(x+6)}
$$
and therefore
$$
f'(x)=\frac{6(x+1)}{5x(x+6)}\sqrt[5]{x}(x+6)=
\frac{6(x+1)}{5\sqrt[5]{x^4}}
$$
However, also your method is sound.
This expression is undefined at $0$ and indeed the function is not differentiable at $0$.
Thus we have two points to take care of: $-1$, where the derivative vanishes, and $0$, where the derivative doesn't exist.
However, the derivative is positive in a neighborhood of $0$ (excluding $0$), so it is increasing both at the left and at the right of $0$. [Note: this is where you're wrong.]
The point $-1$ is a point of minimum, because the derivative is negative at the left and positive at the right of $-1$.
Note that this is confirmed by the fact that
$$
\lim_{x\to-\infty}\sqrt[5]{x}(x+6)=\infty=
\lim_{x\to\infty}\sqrt[5]{x}(x+6)
$$
so the function must have a minimum, being continuous.
Note also that a critical point need not be either a maximum or a minimum: $0$ is a critical point of $f(x)=x^3$, but it's clearly neither a maximum nor a minimum.
