Chapter A3, Problem 8E

To determine

To find: an open interval

Answer to Problem 8E

An open interval is $-\sqrt{3.5}<x<-2,\sqrt{4.5}$

The number is $\delta =0.1213$

Explanation of Solution

Given:

  $f(x)=x^2,L=4,c=-2,\in =0.5$

Calculation:

Step: 2

  $\begin{array}{l}\left|x^2-4\right||<0.5\\ \Rightarrow -0.5<x^2-4<0.5\\ \Rightarrow 3.5<x^2<4.5\\ \Rightarrow -\sqrt{3.5}<x<-2,\sqrt{4.5}\end{array}$

Since $x$ is near $-2,$ we choose the negative square roots

Step: 2

  $\begin{array}{l}|x-(-2)|<\delta \\ \Rightarrow -\delta <x+2<\delta \\ \Rightarrow -2-\delta =-\sqrt{4.5}or-2+\delta =-\sqrt{3.5}\\ \Rightarrow \delta \approx 0.1213or0.1292\end{array}$

Choose the smaller value to satisfy the both conditions

Thus answer is $\delta =0.1213$

Conclusion:

An open interval is $-\sqrt{3.5}<x<-2,\sqrt{4.5}$

The number is $\delta =0.1213$