What are the absolute extrema of #f(x) =x/(x^2-x+1) in[0,3]#?

1 Answer
May 27, 2016

Answer:

Absolute minimum is #0# (at #x=0#) and absolute maximum is #1# (at #x=1#).

Explanation:

#f'(x) = ((1)(x^2-x+1)-(x)(2x-1))/(x^2-x+1)^2 = (1-x^2)/(x^2-x+1)^2#

#f'(x)# is never undefined and is #0# at #x=-1# (which is not in #[0,3]#) and at #x=1#.

Testing the endpoints of the intevral and the critical number in the interval, we find:

#f(0) = 0#
#f(1) = 1#
#f(3) = 3/7#

So, absolute minimum is #0# (at #x=0#) and absolute maximum is #1# (at #x=1#).