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 is \(0.225\).

Step-by-step solution

Solve the given condition

Apply the absolute property, that is, if \(\left| x \right| < a\) then \( - a < x < a\).

Rewrite the given condition \(\left| {{x^2} - 1} \right| < \frac{1}{2}\) as:

\(\begin{aligned} - \frac{1}{2} < {x^2} - 1 < \frac{1}{2}\\ - \frac{1}{2} + 1 < {x^2} < \frac{1}{2} + 1\\\frac{1}{2} < {x^2} < \frac{3}{2}\end{aligned}\)

Thus, \(\frac{1}{2} < {x^2} < \frac{3}{2}\).

Observe the graph

It is observed from the graph and the obtained condition in step 1 that the left part will satisfy the condition \({x^2} = \frac{1}{2}\) and right part will satisfy the condition \({x^2} = \frac{3}{2}\).

Solve the obtained conditions.

\(\begin{aligned}{c}x & = \sqrt {\frac{1}{2}} \\ &= \frac{1}{{\sqrt 2 }}\end{aligned}\)

And,

\(x = \sqrt {\frac{3}{2}} \)

Obtain the value of \(\delta \)

For left side, the condition is \(\left| {x - 1} \right| < \left| {\frac{1}{{\sqrt 2 }} - 1} \right|\) and for right side is \(\left| {x - 1} \right| < \left| {\sqrt {\frac{3}{2}} - 1} \right|\). Simplify both the conditions.

\(\begin{aligned}\left| {x - 1} \right| < \left| {\frac{1}{{\sqrt 2 }} - 1} \right|\\ \approx 0.293\\\left| {x - 1} \right| < 0.293\end{aligned}\)

And,

\(\begin{aligned}\left| {x - 1} \right| < \left| {\sqrt {\frac{3}{2}} - 1} \right|\\ \approx 0.225\\\left| {x - 1} \right| < 0.225\end{aligned}\)

This implies that \(\delta = {\rm{min}}\left\{ {0.293,0.225} \right\}\), that is, \(\delta = 0.225\).

The number \(\delta \) that satisfy the given condition is \(0.225\), that is, \(\left| {x - 1} \right| < 0.225\).