Hello!

Well as the titel says, I need help with a math problem I can't get a grip of. I have thought of this problem for around 1-2 hours and just can't understand it.

Some tips: Use derivate

For which numbers on a doesn't the curve (y = x ^ 3 + ax ^ 2 + x) has any extreme points?

The answer to this question is -√3 < a < +√3

 

Regards!

 

Hello dawoodr, I see you are a new member.

 

The forum has a special homework section where you should have posted this homework question, along with your attempt at solution.

 

Since the hint suggests the derivative, what is the derivative and in particular what sort of polynomial is it?

 

What is the condition that the derivative must satisfy at an extreme point?

This is no homework, it's just a math problem I am having a hard time with.

Edited November 2, 201312 yr by dawoodr

OK so you have an amateur interest in maths. That's good.

 

The derivative is

 

[math]\frac{{dy}}{{dx}} = 3{x^2} + 2ax + 1[/math]

 

How about telling us if you understand this and know what its relationship to extreme points is..

I can't help you if you don't help me understand what you already know.

This is a use of calculus question.

Edited November 2, 201312 yr by studiot

Yes I can show you my solving:

 

[math]\frac{{dy}}{{dx}} = 0[/math]

 

[math]\frac{{dy}}{{dx}} = 3{x^2} + 2ax + 1[/math]

 

[math] 3{x^2} + 2ax + 1 = 0[/math]

 

[math]x = -a/3 +- sqrt({(a/3)^2} - 1/3)[/math]

 

What am I suposed to do now? I know that [math] 3{x^2} + 2ax + 1 = 0[/math] should give you

[math]x = "a number" +- sqrt(a negativ number)[/math]

Edited November 3, 201312 yr by dawoodr

You obviously realize that the derivative is a quadratic (in x)

 

The formulae for the two roots of a quadratic equation are (I have used p, q, and r for the constants to avoid confusion with your equation)

[math]p{x^2} + qx + r = 0[/math]

[math]x = \frac{{ - q + \sqrt {{q^2} - 4pr} }}{{2p}}[/math]

 

[math]x = \frac{{ - q - \sqrt {{q^2} - 4pr} }}{{2p}}[/math]

 

However you do not want these.

 

What you need is the square root part

[math]{\sqrt {{q^2} - 4pr} }[/math]

This is called the discriminant. Since you cannot take a square root of a negative number

 

(q² - 4pr ) must be greater than or equal to zero

 

q² > 4pr

 

Or in the range where

q² < 4pr

 

The quadratic has no solution.

 

Since the quadratic has no solution in this range. there are no extreme points in this range.

 

Can you substitute values from your equation for p. q and r to calculate this range now?

 

It is an easy couple of lines that leads directly to the correct solution.

From this

 

I got it to:

[math]{(a/3)^2} - 1/3 < 0[/math]

[math]a < +-\sqrt{3}[/math]

 

What now? It can't be smaller than both [math]-\sqrt{3}[/math] and [math]+\sqrt{3}[/math]

Edited November 3, 201312 yr by dawoodr

Yes, but once again you do not need to solve the quadratic equation.

 

Just consider the discriminant.

 

My apologies I said the discrinimant was the square root bit, it is only the expression under the root sign

 

(q² - 4pr ) and does not include the sign itself.

 

(q² - 4pr ) > 0

q² > 4pr

 

with p = 3, q = 2a and r = 1

 

4a² > 4.3.1 > 12

a² > 3

For there to be a solution of the quadratic.

 

So there is no solution when

 

a² < 3

[math] - \sqrt 3 [/math][math] < a < [/math][math]\sqrt 3 [/math]

 

The last line in your post 7 is not quite correct since a can be positive or negative, but both have the same value of a².

 

What I have shown above is the correct deduction from the statement that

 

a² < some number

Edited November 3, 201312 yr by studiot