How do you find the critical points for #f'(x)=11+30x+18x^2+2x^3#?

Question

Evelyn Agard · Accepted Answer

The critical numbers for #11+30x+18x^2+2x^3# are:
#-5# and #-1#.

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)#

Aurora Alvarado · Answer

undefined To find the critical points for $ f'(x) = 11 + 30x + 18x^2 + 2x^3 $, we need to find the values of $ x $ where the derivative is equal to zero or undefined.

First, find the derivative of $ f'(x) $ which is $ f''(x) $.

Then, set $ f''(x) = 0 $ and solve for $ x $. These solutions will give you potential critical points.

Additionally, check for values of $ x $ where $ f''(x) $ is undefined, which could also indicate critical points.

After finding these potential critical points, you need to evaluate them to determine if they are indeed critical points by examining the behavior of the function $ f'(x) $ around those points.