Answer
$\delta=\frac{\sqrt5-2}{2}$
Work Step by Step
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-(-1)|\lt\delta\Rightarrow|f(x)-3|\lt0.25$$
Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $2.75$ and $3.25$ (which means $|f(x)-3|\lt0.25$), $x$ must be placed between $-\sqrt5/2$ and $-\sqrt3/2$.
- Calculate the distance from $-1$ to $-\sqrt5/2$ and $-\sqrt3/2$:
$-1-(-\frac{\sqrt5}{2})=-1+\frac{\sqrt5}{2}=\frac{\sqrt5-2}{2}\approx0.118$
$(-\frac{\sqrt3}{2})-(-1)=-\frac{\sqrt3}{2}+1=\frac{2-\sqrt3}{2}\approx0.134$
So we would go with the nearer endpoint, which is $-\sqrt5/2$ and take $\delta$ to be $\frac{\sqrt5-2}{2}$ ( or any smaller positive number is fine).
Therefore, $$0\lt |x-(-1)|\lt\frac{\sqrt5-2}{2}\Rightarrow|f(x)-3|\lt0.25$$
