Question

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

Answer 1

See below.

Explanation:

`abs(x-1) le 1 hArr abs(x-1) +epsilon^2 = 1` with `epsilon in RR` or

`sqrt((x-1)^2) = 1-epsilon^2` or

`(x-1)^2 = (1-e^2)^2 hArr (x-epsilon^2)(x-2+epsilon^2)=0` or

`0 le x le 2`

Now `max abs(x+4)` subjected to `0 le x le 2` is `6` for `x=2`

and

`max abs(x^2-16) = 16` and
`min abs(x^2-16) = 9` for `0 le x le 2` then

`9 le abs(x^2-16) le 16` for `0 le x le 2` then