Answer
$f\left(\frac{1}{16}\right) = -\frac{1}{4}$ is a local minimum value
Preference: The First Derivative Test
Work Step by Step
$f(x) = \sqrt x-\sqrt[4] x$
$f'(x) = \frac{1}{2\sqrt x}-\frac{1}{4\sqrt[4] {x^{3}}} = \frac{2\sqrt[4] {x}-1}{4\sqrt[4] {x^{3}}}$
The domain of the function is $(0,\infty)$.
Use the First Derivative Test:
$2\sqrt[4] {x}-1 \gt 0$
$x \gt \frac{1}{16}$
so $f'(x) \gt 0$ at $x \gt \frac{1}{16}$
and $f'(x) \lt 0$ at $0 \lt x \lt \frac{1}{16}$
Since $f'$ changes from negative to positive at $x$ = $\frac{1}{16}$, $f\left(\frac{1}{16}\right)$ = $-\frac{1}{4}$ is a local minimum value
Use the Second Derivative Test:
$f''(x) = -\frac{1}{4}x^{-\frac{3}{2}}+\frac{3}{16}x^{-\frac{7}{4}}$
$f''\left(\frac{1}{16}\right) = -16+24 = 8 \gt 0$ so
$f\left(\frac{1}{16}\right) = -\frac{1}{4}$ is a local minimum value
Preference: The First Derivative Test may be slightly easier to apply in this case
