Constraints problem

[hajfajv]

Hello

I have a problem where I am to come up with some constraints for a function [math]f(x)=xˆ3+axˆ2+bx+c[/math]so the function has two critical points, [math]x=1[/math] and [math]x=3[/math]We're having about convex and concave functions, first and second derivative tests, mean value therom, and some different ways to find extreme points.

My first thought when seeing this problem is [math]a(x-r1)(x-r2)[/math] but that seems like a bad place to start based on where we're in class.
What I would like is for some guidance on which topic above so I can read more about it, right now I'm just flicking through my book and trying out things but it's very difficult when you have no idea where to look.

 

[Deleted member 4993]

Hello

I have a problem where I am to come up with some constraints for a function [math]f(x)=xˆ3+axˆ2+bx+c[/math]so the function has two critical points, [math]x=1[/math] and [math]x=3[/math]We're having about convex and concave functions, first and second derivative tests, mean value therom, and some different ways to find extreme points.

My first thought when seeing this problem is [math]a(x-r1)(x-r2)[/math] but that seems like a bad place to start based on where we're in class.
What I would like is for some guidance on which topic above so I can read more about it, right now I'm just flicking through my book and trying out things but it's very difficult when you have no idea where to look.

According to your class-notes - what is the definition of "critical point" ? Please be "complete".

 

[JeffM]

Let’s think about polynomials.

The derivative of a polynomial is what kind of function?

The number of real zeroes of a polynomial with real coefficients and degree n cannot exceed what?

 

[Deleted member 4993]

Let’s think about polynomials.

The derivative of a polynomial is what kind of function?

The number of real zeroes of a polynomial with real coefficients and degree n cannot exceed what?

And think about

how are "critical points" related to the "derivatives" of the polynomial?

 

[hajfajv]

According to your class-notes - what is the definition of "critical point" ? Please be "complete".

I believe it is the following:
If a function [imath]f[/imath] is differentiable in an interval [imath]I[/imath] and that [imath]c[/imath] is an interior point of [imath]I[/imath]. If [imath]c[/imath] is maximum or minimum point for [imath]f[/imath] in [imath]I[/imath], then it must be a critical point for [imath]f[/imath], that is, [imath]f'(c)=0[/imath]

or it could be this as they both sound a lot a like to me:
If [imath]f(x)[/imath] has domain [imath]D[/imath], then
(i) [imath]a /isin D[/imath] is a maximum point for [imath]f[/imath], and [imath]f(a)[/imath] is the maximum value, if [imath]f(x) /leq f(a)[/imath]for all [imath]x[/imath] in [imath]D[/imath]
(ii) [imath]b /isin D[/imath] is a minimum point for [imath]f[/imath], and [imath]f(b)[/imath] is the minimum value, if [imath]f(x) /geq f(b)[/imath]for all [imath]x[/imath] in [imath]D[/imath]

 

[hajfajv]

Let’s think about polynomials.

The derivative of a polynomial is what kind of function?

The number of real zeroes of a polynomial with real coefficients and degree n cannot exceed what?

The derivative of a polynomial is a polynomial of one lower degree?

Not sure what you mean be the second question. I understand real zeroes as roots/solutions, real coefficients as the factor in front of "x" or a constant and the degree n as the highest exponent among "x" terms, but I didn't think there would be a limit.

 

[hajfajv]

And think about

how are "critical points" related to the "derivatives" of the polynomial?

Derivatives show the change in rate of the function at a given point, so if you have a critical point it's going to be zero as the tangent is vertical in that point? feel free to correct me.
So if I take the function and take the first derivative and insert one of the roots into the function I can isolate my way out of this question?

 

[hajfajv]

[imath]f'(1)=3+2a+b=0[/imath] and [imath]f'(3)=27+6a+b=0[/imath]
so if I do [imath]3+2a+b=27+6a+b[/imath]
[imath]-24+2a+b=6a+b[/imath]
[imath]-24=6a/2a[/imath]
[imath]-8=a[/imath]

[imath]3+2(-8)+b=0[/imath]
[imath]b=13[/imath]

What am I doing wrong?

 

[BigBeachBanana]

[imath]f'(1)=3+2a+b=0[/imath] and [imath]f'(3)=27+6a+b=0[/imath]
so if I do [imath]3+2a+b=27+6a+b[/imath]
[imath]-24+2a+b=6a+b[/imath]
[imath]-24=6a/2a[/imath]
[imath]-8=a[/imath]

[imath]3+2(-8)+b=0[/imath]
[imath]b=13[/imath]

What am I doing wrong?

Your mistake is equating the 2 equations. Is there a reason why you do that?
Solve for a and b as a system of equations using either elimination or substitution.

 

[Dr.Peterson]

[imath]f'(1)=3+2a+b=0[/imath] and [imath]f'(3)=27+6a+b=0[/imath]
so if I do [imath]3+2a+b=27+6a+b[/imath]
[imath]-24+2a+b=6a+b[/imath]
[imath]-24=6a/2a[/imath]
[imath]-8=a[/imath]

[imath]3+2(-8)+b=0[/imath]
[imath]b=13[/imath]

What am I doing wrong?

Setting the two left-hand sides equal is a valid step in elimination (though not standard); but dividing 6a by 2a is invalid!

You have [imath]-24+2a+b=6a+b[/imath], and after subtracting [imath]b[/imath] from both sides, you have [imath]-24+2a=6a[/imath]. What is the right thing to do to get both terms with [imath]a[/imath] on the same side?

 

[JeffM]

The derivative of a polynomial is a polynomial of one lower degree?

Not sure what you mean be the second question. I understand real zeroes as roots/solutions, real coefficients as the factor in front of "x" or a constant and the degree n as the highest exponent among "x" terms, but I didn't think there would be a limit.

The derivative of a polynomial of degree n is a polynomial of degree n - 1.

The domain of a real polynomial is all real numbers.

The zeroes of a function’s derivative are critical points for that function. Are there others?

So a polynomial that is a polynomial of degree 3 has a derivative that is a polynomial of degree 2.

[math]p \ne 0 \text { and } f(x) = px^3 + qx^2 + sx + t \implies\\ f’(x) = 3px^2 + 2qx + s \implies f’’(x) = 6px + 2q.\\ \therefore f’(x) = 0 \iff 4q^2 \ge 12ps \text { and } f’’(x) = 0 \iff x = - \dfrac{q}{3p}.\\ \therefore 4q^2 = 12ps \implies f’ \left ( - \dfrac{q}{3p} \right ) = 0 = f’’ \left ( - \dfrac{q}{3p} \right ).[/math]
Do NOT memorize the preceding statements. They’re easily derived. Can you do so?

 

[Deleted member 4993]

So if I take the function and take the first derivative and insert one of the roots into the function I can isolate my way out of this question?

My advice to you would be to:

calculate the roots of the derivative of the function and

find the conditions under which you have two real and unique roots.

 

[Dr.Peterson]

Hello

I have a problem where I am to come up with some constraints for a function [imath]f(x)=xˆ3+axˆ2+bx+c[/imath]
so the function has two critical points, [imath]x=1[/imath] and [imath]x=3[/imath]

I think some of us are misreading the problem (as I did initially); it is not merely that there should be two critical points, but specifically that they should be x=1 and x=3. The initial work here, and in the following, are good.

[imath]f'(1)=3+2a+b=0[/imath] and [imath]f'(3)=27+6a+b=0[/imath]

Solving this correctly will give the answer.

 

[Steven G]

I am confused by many of the posts here. The OP said that they need to find the constraints on (I guess) a and b. I feel that there is no reason to solve for a and b.
The OP had the answer when they said 3+2a+b=0 and 27+6a+b=0. I would have written 2a+b = -3 and 6a + b = -27

 

[Dr.Peterson]

I am confused by many of the posts here. The OP said that they need to find the constraints on (I guess) a and b. I feel that there is no reason to solve for a and b.
The OP had the answer when they said 3+2a+b=0 and 27+6a+b=0. I would have written 2a+b = -3 and 6a + b = -27

Actually, we haven't ever been given the actual wording of a problem; this may be part of a larger problem, and not necessarily well understood. Certainly the wording we have contributes to the impression that the goal would be to have two critical points, for which we'd expect an inequality constraint.

Solving for actual values clarifies the "constraint"! (As I read it, the lack of a constraint on c would also be part of the answer!)

@hajfajv, can you show us the entire problem you are working on (both in the original language and a translation, if it is not in English)? That will help us know what you really need.

 

[hajfajv]

Your mistake is equating the 2 equations. Is there a reason why you do that?
Solve for a and b as a system of equations using either elimination or substitution.

I thought I could eliminate b and therefore get a value for a and put that into any of the equations to get a value for b

 

[hajfajv]

Setting the two left-hand sides equal is a valid step in elimination (though not standard); but dividing 6a by 2a is invalid!

You have [imath]-24+2a+b=6a+b[/imath], and after subtracting [imath]b[/imath] from both sides, you have [imath]-24+2a=6a[/imath]. What is the right thing to do to get both terms with [imath]a[/imath] on the same side?

oh I can just subtract 2a, I divided like I was just getting rid of [imath]a[/imath], thanks this should clear this up

 

[hajfajv]

The derivative of a polynomial of degree n is a polynomial of degree n - 1.

The domain of a real polynomial is all real numbers.

The zeroes of a function’s derivative are critical points for that function. Are there others?

So a polynomial that is a polynomial of degree 3 has a derivative that is a polynomial of degree 2.

[math]p \ne 0 \text { and } f(x) = px^3 + qx^2 + sx + t \implies\\ f’(x) = 3px^2 + 2qx + s \implies f’’(x) = 6px + 2q.\\ \therefore f’(x) = 0 \iff 4q^2 \ge 12ps \text { and } f’’(x) = 0 \iff x = - \dfrac{q}{3p}.\\ \therefore 4q^2 = 12ps \implies f’ \left ( - \dfrac{q}{3p} \right ) = 0 = f’’ \left ( - \dfrac{q}{3p} \right ).[/math]
Do NOT memorize the preceding statements. They’re easily derived. Can you do so?

I don't understand line three of the equations, where does the greater then or equalsign come from?

 

[hajfajv]

I think some of us are misreading the problem (as I did initially); it is not merely that there should be two critical points, but specifically that they should be x=1 and x=3. The initial work here, and in the following, are good.

Solving this correctly will give the answer.

you're correct, the problem is giving the roots I just have to find the constraints that satisfy the roots

 

[hajfajv]

I am confused by many of the posts here. The OP said that they need to find the constraints on (I guess) a and b. I feel that there is no reason to solve for a and b.
The OP had the answer when they said 3+2a+b=0 and 27+6a+b=0. I would have written 2a+b = -3 and 6a + b = -27

That is my mistake, the constraints I need to find are of a, b, and c