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