Answer
$(16,32)$
$\delta=7$
Work Step by Step
We are given $f(x)=\sqrt{x-7}$, $L=4$, $c=23$, and $\epsilon=1$.
To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$:
$$\left|\sqrt{x-7}-4 \right| \lt 1\\
-1\lt \sqrt{x-7}-4 \lt 1\\
3\lt \sqrt{x-7} \lt 5\\
9\lt x-7 \lt 25\\
16 \lt x \lt 32.$$
Thus our interval is $(16,32)$.
Now, to find our $\delta$, we note that
$$16 \lt x \lt 32 \implies -7 \lt x-23 \lt 9.$$
Hence for $\delta=7$,
$$0 \lt |x-23| \lt \delta \implies |f(x)-4| \lt \epsilon.$$
