Help with First Derivative Test: Local Maxima and Minima I have a function $f(x)$: $$f\left(x\right)=\frac14x^\left(2/3\right)-x^\left(1/3\right)$$
For which I have found the First derivative: $$f'\left(x\right)=\frac{1}{6x^\left(1/3\right)}-\frac{1}{3x^\left(2/3\right)}$$
Now, the Critical Numbers I have calculated are: $x=0$ and $x=8$. 
However, when I plug them in the first derivative and sketch the number line, I get: $$Local-Maxima=D.N.E.$$ $$Local-Minima=8$$
It appears the Local Minima is wrong, but I don't understand how it is possible. The number line shows me that the function decreases in the interval $(-\infty,0)\cup(0,8)$ and then increases in the interval $(8,\infty)$.
Thanks in advance for the help!
 A: $x=8$ is the location of a local minima and also the absolute minimum. If $f$ is decreasing (going downwards) from the left, hits a critical point, and then starts increasing (going upwards), then this is justification for that critical point being a local minimum. 
More rigorous: $f'(x)=0$ or is undefined at $x=0,8$. Now $f'<0$ on $(-\infty,8)\setminus\{0\}$, so $f$ is decreasing on that interval. $f'(8)=0$ so $x=8$ is the location of a critical point. And $f'>0$ on $(8,\infty)$, so $f$ is increasing on that interval. 
Absolute Minimum: We can reason that $f(8)$ is the absolute minimum by comparing $f(0),\,f(8)$ and the limits at infinity.
\begin{align*}
L &= \lim_{x\to -\infty} f(x) = \infty \\
R &= \lim_{x\to  \infty} f(x) = \infty \\
f(0) &= 0 \\
f(8) &= -1
\end{align*}
So $f(8)<f(0)<L=R$, which implies $f(8)$ is the absolute minimum of $f$.
A: we have $$f''(x)=-\frac{1}{18x^{4/3}}+\frac{2}{9x^{5/3}}$$ and $$f''(8)=\frac{1}{288}>0$$ therefore we have an minimum for $x=8$ with the second derivative test.
