# Zeros of cubic functions

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• Nov 19th 2006, 10:40 AM
topsquark
Quote:

Originally Posted by alexia1huff
I still don't really know how am i to investigate the different roots.

For one root i took (x-1)[cubed] (as you said) and then i found the average root, which is 1 and then the tangent is 0!!! So I don't exactly understand what it means. I guess i'm suppossed to see if my conjecture is true here as well.

i can't find a two roots one. :(

For one real and two complex i took x^3 - 4x^2 + x + 8
but the root that i get from it is not a whole number and i don't really know how to find the complex roots. :confused:

apart from that i have another question, my conjecture is: in cubics the tangent at the average of two roots intersects the curve at the third root. What similar cubics can i use to prove it and they should not be exactly the same, for other proofs you usually have to try functions with different powers, e.g. fractions, negative, but here you can't really do it, so do you have any ideas.
Can you think of any algebric way to prove my conjecture?

There is a way to prove it, but it looks to me like you still really don't have the method down. Let me do another example.

I am going to pick 1, 2, and 3 as roots. The cubic polynomial that does that is:
$\displaystyle (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6$

I am going to show that the line tangent to the cubic at $\displaystyle x = \frac{1 + 2}{2} = \frac{3}{2}$ passes through third root of the cubic, (3, 0).

The slope of the required line is:
$\displaystyle m = \frac{d}{dx}(x^3 - 6x^2 + 11x - 6) = 3x^2 - 12x + 11$
evaluated at x = 3/2:
$\displaystyle m = -\frac{1}{4}$

We know this line is to be tangent to the cubic at x = 3/2, so we need to find the correct y-intecept of this line:
$\displaystyle y = mx + b$

So we need to know which y corresponds to the cubic with a value of x = 3/2:
$\displaystyle y = x^3 - 6x^2 + 11x - 6$

So $\displaystyle y(3/2) = \frac{3}{8}$

Plugging this into the line equation:
$\displaystyle \frac{3}{8} = -\frac{1}{4} \cdot \frac{3}{2} + b$

So $\displaystyle b = \frac{3}{4}$.

So our tangent line is
$\displaystyle y = -\frac{1}{4}x + \frac{3}{4}$

Now, this presumably crosses the x-axis at the same point as the third root of the cubic: (3, 0). Let's find out:

$\displaystyle 0 = -\frac{1}{4}x + \frac{3}{4}$

$\displaystyle 0 = -x + 3$

$\displaystyle x = 3$

So the line does indeed intersect the point (3, 0). The graph below gives a visual indication of this.

-Dan
• Nov 19th 2006, 10:53 AM
topsquark
Quote:

Originally Posted by alexia1huff
For one root i took (x-1)[cubed] (as you said) and then i found the average root, which is 1 and then the tangent is 0!!! So I don't exactly understand what it means. I guess i'm suppossed to see if my conjecture is true here as well.

For this case what you would find is that the "tangent line" is y = 0, which passes through the point (1, 0). This line isn't really a tangent, but it does still validate the theorem.

Quote:

Originally Posted by alexia1huff
i can't find a two roots one. :(

For one real and two complex i took x^3 - 4x^2 + x + 8
but the root that i get from it is not a whole number and i don't really know how to find the complex roots. :confused:

What are you talking about? YOU decide the roots! Nobody is telling you to take an arbitrary cubic and do the theorem. Make up one where you know the roots and use that.

Quote:

Originally Posted by alexia1huff
have any ideas.
Can you think of any algebric way to prove my conjecture?

Warning: This is going to be a bit of a messy one.
The conjecture can be proven like this: (This follows the work in my previous post.)
Choose 3 roots, p, q, and r. (For the moment I am considering them all to be real numbers, the proof for complex roots is similar.)

So the cubic is:
$\displaystyle y = (x - p)(x - q)(x - r) = x^3 - (p+q+r)x^2 +(pq + pr + qr)x - pqr$

You wish to find the line tangent to this cubic at the point $\displaystyle x = \frac{p+q}{2}$. It suffices to choose this point as any other choice of roots to average merely shuffles the p, q, r's around.

The slope of this line will be the first derivative of y evaluated at the average x value above.

Now find out what the y value of the point on the tangent line is. (Evaluate y for the average x value.)

Now you need the intercept of the tangent line. So y = mx + b, solve for b.

Now you need to show that this line intersects the point (r, 0). So solve 0 = mx + b for x. The solution should be x = r.

-Dan
Show 40 post(s) from this thread on one page
Page 2 of 2 First 12