Identifying and Classifying Stationary Points for f(x) = x^3 + ax^2 + 5x + 2

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Homework Help Overview

The problem involves the function f(x) = x^3 + ax^2 + 5x + 2, which is stated to have two stationary points, one of which is at x = 2. The tasks include finding the other stationary point and classifying the stationary points using the second derivative test.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivative f'(x) and its role in identifying stationary points. There is confusion regarding the correct application of conditions for stationary points, with some participants questioning the use of f(x) instead of f'(x) in their calculations. Various attempts to solve for 'a' and find the other stationary point are presented, including different approaches to substituting values into the equations.

Discussion Status

The discussion is ongoing, with participants exploring different methods to find the stationary points and clarify the conditions necessary for their identification. Some guidance has been offered regarding the need to substitute x = 2 into the entire equation, indicating a productive direction in the conversation.

Contextual Notes

There appears to be some confusion regarding the definitions and conditions for stationary points, as well as the correct application of derivatives in the context of this problem.

Carl_M
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Homework Statement



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

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

For a stationary point, f'(x) = 0, not f(x) = 0.
 
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:
You have to substitute x=2 in the entire equation, not just one variable.
 

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