Question

Use the given graph of \(f\left( x \right) = {x^2}\)to find a number \(\delta \)such that if \(\left| {x - 1} \right| < \delta \)then\(\left| {{x^2} - 1} \right| < \frac{1}{2}\).

Short answer

The number \(\delta \) is \(0.225\) such that if \(\left| {x - 1} \right| < \delta \)then\(\left| {{x^2} - 1} \right| < \frac{1}{2}\).

Step-by-step solution

Describe the given information

The given function is as follows;

\(f\left( x \right) = {x^2}\)

Consider the given graph.

Find the number \(\delta \)

To determine the number \(\delta \) such that \(\left| {x - 1} \right| < \delta \)then \(\left| {{x^2} - 1} \right| < \frac{1}{2}\), use the given graph of the function.

Expand the inequality\(\left| {x - 1} \right| < \delta \).

\(\begin{array}{c} - \delta < x - 1 < \delta \\ - \delta + 1 < x < \delta + 1\end{array}\)

Therefore, for every\(x \in \left( {1 - \delta ,\;1 + \delta } \right)\), the corresponding values of \(f\left( x \right)\) lie on\(\left( {0.5,\;1.5} \right)\).

Locate the numbers \(1 - \delta \) and \(\delta + 1\) on the graph of \(f\left( x \right) = {x^2}\).

From the graph, the value of \(f\left( x \right) = \sqrt x \) corresponding to \(x = 1 - \delta \)is \(0.5\). So, \(f\left( {1 - \delta } \right) = 0.5\).

\(\begin{array}{c}{\left( {1 - \delta } \right)^2} = 0.5\\1 - \delta = \sqrt {0.5} \\\delta = 1 - \sqrt {0.5} \\ = 0.293\end{array}\)

From the graph, the value of \(f\left( x \right) = \sqrt x \) corresponding to \(x = 1 + \delta \)is \(1.5\). So, \(f\left( {1 + \delta } \right) = 1.5\).

\(\begin{array}{c}{\left( {1 + \delta } \right)^2} = 1.5\\1 + \delta = \sqrt {1.5} \\\delta = \sqrt {1.5} - 1\\ = 0.225\end{array}\)

The required number \(\delta \) will be a minimum of two numbers \(0.293\)and\(0.225\).

\(\begin{array}{c}\delta = \min \left( {0.225,\;0.293} \right)\\ = 0.225\end{array}\)

Therefore, the number \(\delta \) is \(0.225\) such that if \(\left| {x - 1} \right| < \delta \)then\(\left| {{x^2} - 1} \right| < \frac{1}{2}\).