I'm not sure why you called the function #f'(x)#. Since we need to take the derivative to find the critical numbers, I'll refer to the function as #g(x)#.
The critical numbers for a function #g# are the numbers in the domain of #g# at which the derivative is either #0# or does not exist. Some people use "critical point" to mean the same thing, others use it to mean a point on the graph (so it has 2 coordinates).
#g(x) = 11+30x+18x^2+2x^3#
#g'(x) = 30+36x+6x^2#
This is never undefined, so we need only find the zeros:
#6x^2+36x+30 = 0#
#6(x^2+6x+5)=0#
#6(x+5)(x+1) = 0#
#x=-5# or #x=-1#
The critical numbers for the function: #11+30x+18x^2+2x^3# are #-5# and #-1#.
If you wish to find the #y# values, you can do so.
At #x=-5#, we get:
#11+30(-5)+18(-5)^2+2(-5)^3#
(I prefer to do arithmetic with smaller numbers. Using #30 = 6*5#, we can get some multiples of #25#. So I'll regroup and use the distributive property.)
#11-6(25)+18(25)-10(25) = 11+18(25)-16(25)#
# = 11+[18-16] (25) = 11+2(25)#
# = 11 + 50 = 61#
At #x=-1#, we get:
#11-30+18-2 = 29-32 = -3#
So, if you use "critical points" to mean points on the graph (rather than points in the domain), then they will be:
#(-5, 61)# and #(-1, -3)#
