How do you find all the extrema for #f(x) = x/(x^2+x+1)# on [-2,2]?

Question

1997 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-10T13:43:24

Relative Max: #1/3# (at #1#), Relative min #-1# (at#-1#)
Absolute max and min are the same as the relative.

#f(x) = x/(x^2+x+1)# on [-2,2]

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

#f'(x)# is never undefined and is #0# at #x = +-1#

On #[-2,-1)#, we have #f'(x) <0#
on #(-1,1)#, we have #f'(x) > 0#

So #f(-1) = -1# is a relative minumum.

On #(-1,1)#, we have #f'(x) >0#
on #(1,2]#, we have #f'(x) < 0#

So #f(1) = 1/3# is a relative minumum.

#f(-2) = -2/3 > -1# and #f(2) = 2/7 < 1/3# so the relative extrema are also absolute on the interval.