Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 6

We need to use a graph to find a number $\delta$ such tha if $|x - 1| < \delta $ then $\displaystyle \left| \frac{2x}{x^2 + 4} - 0.4 \right| < 0.1$








First, we will get the values of $x$ that intersect at the given curve to their corresponding $y$ values. Let $x_L$ and $x_R$
are the values of $x$ from the left and right of 1 respectively.

$
\begin{equation}
\begin{aligned}
y & = \frac{2x_L}{(x_L)^2+4} \\
0.3 & = \frac{2x_L}{(x_L)^2+4} \\
0.3 (x_L)^2 + 1.2 & = 2x_L \\
0.3 x_L^2 - 2x_L + 1.2 & = 0 \\
\end{aligned}
\end{equation}
$


Using quadratic formula,

$
\begin{equation}
\begin{aligned}
& x_{L(1,2)} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\
\\
& x_{L(1,2)} = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(0.3)(1.2)}}{2(0.3)}\\
\\
& x_{L1} = 6 \text{ and } x_{L2} = \frac{2}{3}
\end{aligned}
\end{equation}
$


Evaluating $x$ to the right of 1,

$
\begin{equation}
\begin{aligned}
y & = \frac{2x_R}{(x_R)^2+4}\\
0.5 & = \frac{2x_R}{(x_R)^2+4}\\
0.5 (x_R)^2 + 2 & = 2x_R\\
0.5 x_R^2 - 2x_R + 2 & = 0\\
\end{aligned}
\end{equation}
$


Using quadratic formula,
$ \qquad x_R = 2$

Now, we can determine the value of $\delta$ by checking the values of $x$ that would give a smaller distance to 1.


$
\begin{equation}
\begin{aligned}
1 - x_{L2} = 1 - \frac{2}{3}& = \frac{1}{3}\\
x_{R} - 1 = 2 - 1 & = 1
\end{aligned}
\end{equation}
$


Hence,
$\displaystyle \quad \delta \leq \frac{1}{3}$

This means that by keeping $x$ within $\displaystyle \frac{1}{3}$ of $1$, we are able to keep $f(x)$ within $0.1$ of $0.4$.

Although we chose $\delta =\displaystyle \frac{1}{3}$, any smaller positive value of $\delta$ would also have work.