Answer
$\delta=1.44$
(or any smaller positive number)
Work Step by Step
First, we address the question marks on the x-axis.
They stand for the numbers whose function values are 1.6 and 2.4
$\sqrt{x}=1.6 \ \ \Rightarrow\ \ x=1.6^{2}=2.56$
$\sqrt{x}=2.4 \ \ \Rightarrow\ \ x=2.4^{2}=5.76$
So, the interval (to which x=4 belongs) is $(2.56, 5.76)$
or, rewritten as $I=(4-1.44, 4+1.76).$
From the graph, $x\in I \ \ \Rightarrow\ \ |f(x)-2| < 0.4$.
(all values of $x\in I$ have f(x) within $\pm 0.4 $ of 2)
We take a subset from $I$, one that we can write as$ (4-\delta,4+\delta)$,
such as
$I_{1}=(4-1.44,4+1.44)$
For $x\in I_{1}\ \ \Rightarrow\ \ x\in I\ \ \Rightarrow\ \ |f(x)-2| < 0.4$.
So for $\epsilon=0.4, $we have found $\delta=1.44$, such that
$ 0 < |x-4| < \delta$, then $|f(x)-2| < \epsilon$.
