# Finding Stationary Points

1. Feb 2, 2010

### Carl_M

1. The problem statement, all variables and given/known data

f(x) = x^3 +ax^2 +5x +2 and has two stationary points, one of which is x = 2. a) Find the other stationary point.
b) classify the stationary points using the second derivative test

3. The attempt at a solution

f '(x) = 2ax + 3x^2 +5
f ''(x) = 2a + 6x

0 = (2)^3 + a(2^2) +5(2) +2
0 = 8 + 4a + 10 +2
0 = 20 + 4a
-4a = 20, a = -5

What am I supposed to do to get a stationary point?

Last edited: Feb 2, 2010
2. Feb 2, 2010

0 = (2)^3 + a(2^2) +5(2) +2

For a stationary point, f'(x) = 0, not f(x) = 0.

3. Feb 2, 2010

### Carl_M

f ' (x) = 0
So that would be?
2ax + 3x^2 + 5 = 0
How would I find a?

Would I use 0 = (2)^3 + a(2^2) +5(2) +2 to get a = -5

0 = 2(-5)x + 3x^2 +5
0 = -10x + 3x^2 +5
0 = 3x^2 - 10x + 5
x = (5 +sqrt(10))/3 and (5 - sqrt(10))/3 ?

or

0 = ( 2a(2) + 3(2)^2 +5
0 = 4a + 12 + 5
0 = 4a +17
-4a = 17
a = -17 / 4

substitute -17/4 into 2ax + 3x^2 + 5 = 0?
0 = 2(-17/4)x + 3x^2 + 5 = 0
0 = 3x^2 - 8.5x + 5 = 0
x = 2 , 5/6

the second derivative?

Last edited: Feb 2, 2010
4. Feb 2, 2010

### Bohrok

You have to substitute x=2 in the entire equation, not just one variable.