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

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Answer 1 · 2017-10-22T17:52:36

See below.

$abs(x-1) le 1 hArr abs(x-1) +epsilon^2 = 1$ 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

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