Answer
$(\frac{1}{16},-\frac{1}{4})$ is the local minimum of $f$. There is no local maximum.
Work Step by Step
$$f(x)=\sqrt x-\sqrt[4]x$$ $$f(x)=x^{1/2}-x^{1/4}$$
1) Find $f'(x)$ and $f''(x)$
$$f'(x)=\frac{1}{2}x^{-1/2}-\frac{1}{4}x^{-3/4}$$ $$f'(x)=\frac{1}{2}x^{-3/4}x^{1/4}-\frac{1}{4}x^{-3/4}$$ $$f'(x)=\frac{1}{2}x^{-3/4}(x^{1/4}-\frac{1}{2})$$ $$f'(x)=\frac{\sqrt[4]x-\frac{1}{2}}{2\sqrt[4]{x^3}}$$
$$f''(x)=\frac{1}{2}\frac{-1}{2}x^{-3/2}-\frac{1}{4}\frac{-3}{4}x^{-7/4}$$ $$f''(x)=-\frac{1}{4\sqrt{x^3}}+\frac{3}{16\sqrt[4]{x^7}}$$
2) The First Derivative Method
Since there exists $\sqrt[4]x$ and $\sqrt[4]{x^3}$ in $f'$, the domain of $f'$ here is $[0,+\infty)$. In other words, we would not consider the interval $(-\infty,0)$.
Also, we notice that the denominator $2\sqrt[4]{x^3}\gt0$ for all $x\in(0,+\infty)$.
Now we need to build a sign table for $f'(x)$. Considering all the facts above, we only need to take one point where the change of signs happen, which is $\sqrt[4]x=\frac{1}{2}$, or $x=\frac{1}{16}$.
The table has been built in the image below.
As we observe the table, $f'$ only changes from negative to positive at $x=\frac{1}{16}$, so $f$ has a local minimum at $x=\frac{1}{16}$. $f'$ does not change from positive to negative anytime.
$$f(\frac{1}{16})=\sqrt{\frac{1}{16}}-\sqrt[4]{\frac{1}{16}}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$$
Therefore, $(\frac{1}{16},-\frac{1}{4})$ is the local minimum of $f$. There is no local maximum.
3) The Second Derivative Method
Find $f'(x)=0$ $$\frac{\sqrt[4]x-\frac{1}{2}}{2\sqrt[4]{x^3}}=0$$ $$\sqrt[4]x-\frac{1}{2}=0$$ $$\sqrt[4]x=\frac{1}{2}$$ $$x=\frac{1}{16}$$
Now the job is find out sign of $f''(\frac{1}{16})$.
$$f''(\frac{1}{16})=-\frac{1}{4\sqrt{\frac{1}{16^3}}}+\frac{3}{16\sqrt[4]{\frac{1}{16^7}}}=-16+24=8$$
$f''(\frac{1}{16})=8\gt0$, so $f$ has a local minimum at $x=\frac{1}{16}$, which is $(\frac{1}{16},-\frac{1}{4})$.
$f$ has no local maximum.
4) Comparison
Both tests give out the same result. However, for me, I find the calculations in the First Test less daunting than those in the Second Derivative Test. So I still prefer the First Derivative Test.
