HELP! ASAP!!

[sjmaddox]

given f(x)=ax³−15x²+ bx+ 15.
When XE(3,5), f(x)<0

- determine the values of a and b
- determine all other intervals where f(x)<0

How do I find these values?

 

[MarkFL]

Hello, and welcome to FMH!

Do you mean:

[MATH]f(x)=ax^3-15x^2+bx+15[/MATH] ?

 

[Steven G]

If you would like to be helped then please write a clear post.

Did you mean to have x and X? What does XE(3,5) mean??

In defining f(x) you had 15x^2X. Did you mean that?

Also, please read the guidelines so that you can receive help.

 

[sjmaddox]

Hello, and welcome to FMH!

Do you mean:

[MATH]f(x)=ax^3-15x^2+bx+15[/MATH] ?

Yes, I just figured out how to work it.

 

[sjmaddox]

If you would like to be helped then please write a clear post.

Did you mean to have x and X? What does XE(3,5) mean??

In defining f(x) you had 15x^2X. Did you mean that?

Also, please read the guidelines so that you can receive help.

Yes, x is one of the values. And X is because it is an inequality. So it is apart of the "therefore" statement (x is an element of...)

I was extremely desperate and did not expect an answer in all honesty. This is how the question is worded on my page. I would be extremely thankful if you were able to help!

 

[MarkFL]

I would begin by observing that:

[MATH]f(3)=0[/MATH]
[MATH]f(5)=0[/MATH]
This will give us two equations in two unknowns. Can you proceed?

 

[sjmaddox]

I would begin by observing that:

[MATH]f(3)=0[/MATH]
[MATH]f(5)=0[/MATH]
This will give us two equations in two unknowns. Can you proceed?

yes, I figured that because it was an inequality that XE(3,5) were x-ints and used the remainder theorem from there. but I'm just stuck at the stage before identifying the variables...unless that's wrong....

 

[Dr.Peterson]

Presumably the problem is:

Given [MATH]f(x)=ax^3-15x^2+bx+15[/MATH], when [MATH]x\in (3,5), f(x)\lt0[/MATH].​

I think we have to assume it means that (3, 5) is one of the entire intervals on which f(x) is negative (not just a subset of such an interval!), so that f(x) = 0 at 3 and 5. That allows you to find a and b, and then you can find the other zero of f.

 

[MarkFL]

yes, I figured that because it was an inequality that XE(3,5) were x-ints and used the remainder theorem from there. but I'm just stuck at the stage before identifying the variables...unless that's wrong....

This allows us to write:

[MATH]a(3)^3-15(3)^2+b(3)+15=0[/MATH]
[MATH]a(5)^3-15(5)^2+b(5)+15=0[/MATH]
This is a linear system of equations in the parameters \(a\) and \(b\)...can you solve it?

 

[sjmaddox]

Presumably the problem is:

Given [MATH]f(x)=ax^3-15x^2+bx+15[/MATH], when [MATH]x\in (3,5), f(x)\lt0[/MATH].​

I think we have to assume it means that (3, 5) is one of the entire intervals on which f(x) is negative (not just a subset of such an interval!), so that f(x) = 0 at 3 and 5. That allows you to find a and b, and then you can find the other zero of f.

Thank you!

Question/concern: I tried setting f[MATH]x[/MATH] to 0 with 3 and 5, and I tried using elimination to get a and b, but i kept getting "unfriendly" fractions, is that expected for this question?

 

[MarkFL]

I'm getting integral values.

 

[sjmaddox]

I'm getting integral values.

I don't know what those are, sorry. Can you clarify?

 

[MarkFL]

I don't know what those are, sorry. Can you clarify?

I'm getting values that are integers. Can you write a simplified form of the system of equations and we can go from there?

 

[sjmaddox]

I'm getting values that are integers. Can you write a simplified form of the system of equations and we can go from there?

ahahaha i'm having a blonde moment right now...

I got 27a + 3b -120 = 0
and 125a + 5b -360 = 0

does that seem right on your end?

 

[MarkFL]

ahahaha i'm having a blonde moment right now...

I got 27a + 3b -120 = 0
and 125a + 5b -360 = 0

does that seem right on your end?

Yes, those look good...I would even consider dividing the first equation by 3 and the second by -5 to get

[MATH]9a+b-40=0[/MATH]
[MATH]-25a-b+72=0[/MATH]
Now add the equations...what do you get?

 

[sjmaddox]

Yes, those look good...I would even consider dividing the first equation by 3 and the second by -5 to get

[MATH]9a+b-40=0[/MATH]
[MATH]-25a-b+72=0[/MATH]
Now add the equations...what do you get?

Okay, I got a = 2 and b = 22. yes?

 

[MarkFL]

Okay, I got a = 2 and b = 22. yes?

Yes, that's what I get too. Can you now find the third zero of the cubic function?

 

[sjmaddox]

Yes, that's what I get too. Can you now find the third zero of the cubic function?

so you get f(x) = 2³-15²+22+15

and factor?

 

[sjmaddox]

Yes, that's what I get too. Can you now find the third zero of the cubic function?

whoops, forgot the x's on that last equation

 

[MarkFL]

Yes:

[MATH]f(x)=2x^3-15x^2+22x+15[/MATH]
You could factor, or use division knowing two of the zeroes.