use the given graph of $f(x)=x^{2}$ to find a number $delta$ such that if $|x - 1|<delta$ then…

Explanation:

Step1: Rewrite the inequality $|x^{2}-1| < \frac{1}{2}$

We know that $|x^{2}-1|=(x - 1)(x + 1)$. The inequality $|x^{2}-1|<\frac{1}{2}$ is equivalent to $-\frac{1}{2}<x^{2}-1<\frac{1}{2}$, which simplifies to $\frac{1}{2}<x^{2}<\frac{3}{2}$.

Step2: Solve for $x$ in $\frac{1}{2}<x^{2}<\frac{3}{2}$

Taking square - roots, we get $\sqrt{\frac{1}{2}}<|x|<\sqrt{\frac{3}{2}}$. Since we are interested in values of $x$ close to $1$, for $x>0$, we have $\frac{\sqrt{2}}{2}<x<\frac{\sqrt{6}}{2}$.

Step3: Find the distance from $x = 1$

We want to find $\delta$ such that if $|x - 1|<\delta$, then $|x^{2}-1|<\frac{1}{2}$. First, find the distance from $x = 1$ to the left - hand and right - hand bounds. The distance from $x = 1$ to $x=\frac{\sqrt{2}}{2}\approx0.707$ is $1-\frac{\sqrt{2}}{2}\approx1 - 0.707 = 0.293$. The distance from $x = 1$ to $x=\frac{\sqrt{6}}{2}\approx1.225$ is $\frac{\sqrt{6}}{2}-1\approx1.225 - 1=0.225$. We take the smaller of these two distances.

Answer:

$0.225$