# Minimum and maximum values for abs(x^2-16) given abs(x-1) le 1 ?

Oct 22, 2017

See below.

#### Explanation:

$\left\mid x - 1 \right\mid \le 1 \Leftrightarrow \left\mid x - 1 \right\mid + {\epsilon}^{2} = 1$ with $\epsilon \in \mathbb{R}$ or

$\sqrt{{\left(x - 1\right)}^{2}} = 1 - {\epsilon}^{2}$ or

${\left(x - 1\right)}^{2} = {\left(1 - {e}^{2}\right)}^{2} \Leftrightarrow \left(x - {\epsilon}^{2}\right) \left(x - 2 + {\epsilon}^{2}\right) = 0$ or

$0 \le x \le 2$

Now $\max \left\mid x + 4 \right\mid$ subjected to $0 \le x \le 2$ is $6$ for $x = 2$

and

$\max \left\mid {x}^{2} - 16 \right\mid = 16$ and
$\min \left\mid {x}^{2} - 16 \right\mid = 9$ for $0 \le x \le 2$ then

$9 \le \left\mid {x}^{2} - 16 \right\mid \le 16$ for $0 \le x \le 2$ then