Answer
For $h$ near $16$ ,
$$
f(h) \approx \sqrt{16}+\frac{1}{2 \sqrt{16}}(h-16)=4+\frac{1}{8}(h-16)
$$
$f(h) \geq 0$ if $h>-16$
Work Step by Step
For $h$ near $16$ ,
$$
f(h) \approx \sqrt{16}+\frac{1}{2 \sqrt{16}}(h-16)=4+\frac{1}{8}(h-16)
$$
$f(h) \geq 0$ if $h>-16$
