Answer
$(3,15)$
$\delta=5$
Work Step by Step
We are given $f(x)=\sqrt{19-x}$, $L=3$, $c=10$, and $\epsilon=1$.
To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$:
$$\left|\sqrt{19-x}-3 \right| \lt 1\\
-1 \lt \sqrt{19-x}-3 \lt 1\\
2\lt \sqrt{19-x} \lt 4\\
4\lt 19-x \lt 16\\
-15 \lt -x \lt -3\\
15\gt x \gt 3\\
3 \lt x \lt 15.$$
Thus our interval is $(3,15)$.
Now, to find our $\delta$, we note that
$$3 \lt x \lt 15 \implies -7\lt x-10 \lt 5.$$
Hence for $\delta=5$,
$$0 \lt |x-10| \lt \delta \implies |f(x)-3| \lt \epsilon.$$
