The critical values will occur when the derivative is #0# or undefined.
#f'(x) = (1(x^2+ x +1) - (x + 1)(2x + 1))/(x^2 + x + 1)^2#
#f'(x) = (x^2 + x + 1 - (2x^2 + 3x + 1))/(x^2 + x+ 1)^2#
#f'(x) = (-x^2 - 2x)/(x^2 + x + 1)^2#
Set the derivative to #0# and solve. Also, find the vertical asymptotes (where the function is undefined).
#0 = (-x^2 - 2x)/(x^2 + x+ 1)^2#
#0 = -x^2 - 2x#
#0 = -x(x + 2)#
#x = 0 and -2#
For V.A:
#(x^2 + x + 1)^2 = 0#
#x^2 + x + 1 = 0#
#x = (-1 +- sqrt(1^2 - 4 xx 1 xx 1))/(2 xx 1)#
#x = (-1 +- sqrt(-3))/2#
#:.#There are no vertical asymptotes.
Hence, the critical numbers are #x= 0# and #x= -2#.
Hopefully this helps!
