Chapter D, Problem 6E

To determine

To calculate:

Value of $\delta$ such that if $\left|x-1\right|<\delta$ then $\left|\frac{2x}{x^2+4}-0.4\right|<0.1$

Answer to Problem 6E

The value of number $\delta$ is $0<\delta <0.34$ .

Explanation of Solution

Given information:

Condition given: if $\left|x-1\right|<\delta$ then $\left|\frac{2x}{x^2+4}-0.4\right|<0.1$

Calculation:

Consider the given function,

  $f\left(x\right)=\left|\frac{2x}{x^2+4}-0.4\right|$

Then we have to find $\delta$ such that,

if $\left|x-1\right|<\delta$ then $\left|\frac{2x}{x^2+4}-0.4\right|<0.1$

now,

  $\left|\frac{2x}{x^2+4}-0.4\right|<0.1$

Or $-0.1<\left|\frac{2x}{x^2+4}-0.4\right|<0.1$

Or $0.3<\left|\frac{2x}{x^2+4}-0.4\right|<0.5$

The graph of $f\left(x\right)=\left|\frac{2x}{x^2+4}-0.4\right|$ is shown as:

  

Here drawn a horizontal line $y=0.3,y=0.5$ intersect the graph of the function $f\left(x\right)=\frac{2x}{x^2+4}$ at $\left(0.66,0.3\right)$

At $y=0.3,y=0.5$ the value of $x$ is $0.66,2$ .

In case it is restrict the value of $x$ between $\left(0.66,2\right)$ , then the graph of $f\left(x\right)=\frac{2x}{x^2+4}$ lies between the lines $y=0.3$ and $y=0.5$ .

The distance of $x=1$ from $\left(0.66,0.3\right)$ and $\left(2,0.5\right)$ are

  $\begin{array}{l}1-0.66=0.34\\ 2-1=1\end{array}$

The interval $\left(0.66,2\right)$ is symmetric about $x=1$ .

Nearest point from $x=1$ is $\left(0.66,0.3\right)$ so nearest distance is $\left(0.876,1.2\right)$ .

Nearest distance is $0.34$ .

Thus, $\left|x-1\right|<\delta$ then $\left|\frac{2x}{x^2+4}-0.4\right|<0.1$

Hence the value of number $\delta$ is $0<\delta <0.34$ .