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

1 Answer
Jul 27, 2015

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

Explanation:

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)