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

1 Answer
May 27, 2016

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).