cubic function f(x) = 2x^3 + 6x^2 - 4.5x -13.5

[alexia1huff]

Could you please help me with the following?

Suppose you have the cubic function f(x) = 2x^3 + 6x^2 - 4.5x -13.5.

1) Find the roots and confirm them by Remainder Theorem.

2) Taking two roots at a time, find the equations of the tangent lines to the average of two of the three roots?

3) Find where the tangent lines at the average of the two roots intersect the curve again.

4) State a conjecture concerning the roots of the cubic and tangent lines at the average value of the roots. Proove it and investigatene root, two roots and, one real and two complex roots.

1) I found the roots from the graph -- namely (1.5, 0), (-3, 0) and (-1.5, 0) -- but could someone tell me in steps how to factorise it to find the roots in the form
a(x - b)(x - c)(x - d)? Or any other way of finding the roots...?

When I'm supposed to prove the roots by using the Remainder Thorem, do I divide the function by each of the roots, one at a time?

2) I guess you take ((x - b) + (x - c))/2 for the average of roots, and I think you're suppposed to use differentiation to find the tangents. That is, I think I'm supposed to differentiate the function and then put the average of the roots in it. If so, then what do you do with the number you get from this calculation? In what form should the tangent be written?

4) I'm also suppossed to see some pattern in my calculations. Am I supposed to look at each pair of roots? Because I guess that it won't be all the pair that will intersect the curve again.

:evil:

 

[galactus]

Re: cubic functions

Could you please help me with the following:

Cubic function f(x)=\(\displaystyle 2x^{3} + 6x^{2} - \frac{9}{2}x -\frac{27}{2}\).

1) Find the roots and confirm them by remainder theorem.

You can find the roots of the cubic by the Remainder theorem as mentioned.

The trick is to look at the constant term and leading coefficient.

Find the common factors. It's too much to go into here. I would suggest a good text or an online search for in-depth.

You can just check by trial and error after you find the candidates.

-3 is a root.

Divide your cubic by (x+3):

First multiply the cubic by 2 to get rid of the fractions:

\(\displaystyle \L\\\frac{4x^{3}+12x^{2}-9x-27}{x+3}=4x^{2}-9\)

Now, all you have is a quadratic. Find it's roots and you have them all.

 

[alexia1huff]

I can't do it by trial and error because i have to show my working out and the average is not ((x-a)+(x-b))/2, but a+b/2.

 

[marcmtlca]

Hi, so if you want to write up a reasoning, you would say the following:

1) start by rewriting the equation as galactus did
2) notice that all the coefficients are integers
3) conclude that by the remainder theorem, any real roots must be rational and expressible as:

\(\displaystyle \frac{p}{q}\)

where \(\displaystyle q \in \{ \pm 1, \pm 2, \pm 4 \}\)
and
where \(\displaystyle p \in \{\pm 1, \pm 3, \pm 9 \pm 27 \}\)

you can write some of these out until you stumble upon one which works

4) divide the polynomial by the \(\displaystyle x-\frac{p}{q}\) and then you are left with an degree-2 problem.

 

[skeeter]

f(x) = 2x^3 + 6x^2 - 4.5x - 13.5

f(x) = 2x^2(x + 3) - 4.5(x + 3)

f(x) = (x + 3)(2x^2 - 4.5)

f(x) = 2(x + 3)(x - 1.5)(x + 1.5)

 

[alexia1huff]

Thanks for the factorisation. Now I got as far as 4., but I need help again
how do i proove algebrically my conjecture, which will be: In cubics the tangent at the average of two roots intersects the curve again at the third root.

Also what examples can i use to proove it? Is it possible to take x^3?

How do I investigate the roots? I guess I have to show if the conjecture works for these different roots also, but wouldn't that be for examples to proove.

For one root i'm using, (x-1)^3. The tangent here is 0. So i don't really get what to do.
I can't find a cubic with 2 roots.
For one real root and two complex I'm using x^3-4x^2+x+8