14th Degree Polynomial Help!

**Jeka** · October 1st 2012, 09:19 PM

I need help find all the turning points of this function:

f(x) = x^(14)-12x^(13)-21x^(12)+744x^(11)-1326x^(10)-16800x^(9)+56862x^(8)+153768x^(7)-817347x^(6)-253652x^(5)+4829511x^(4)-3026448x^(3)-9283680x^(2)+5702400x+7776000

the 14th degree polynomial is the expanded one. Here is the original one:

f(x) = (x+1)^(2)(x-5)^(3)(x+4)^(4)(x-3)^(5)

What i want to know is that, how can i find the turning points. I want to know how you can find all the turning points of this function. If you could show, can you please list all the working out, please!

Thanks!

**MaxJasper** · October 1st 2012, 10:16 PM

**Jeka** · October 2nd 2012, 04:54 PM

What does the graph show?
I don't really need the graph.
What I want to know is that, how can i find all the turning points of the function

And by the way, what program did you use to graph it?

**richard1234** · October 2nd 2012, 06:31 PM

Originally Posted by Jeka

What does the graph show?
I don't really need the graph.
What I want to know is that, how can i find all the turning points of the function

And by the way, what program did you use to graph it?

Actually the graph turns out to be very useful. Turning points occur when the derivative is zero, and changes signs (from + to -, or vice versa). The dashed line represents the derivative of $f(x)$ (denoted $f'(x)$ or $\frac{d}{dx} f(x)$ ) and intersects the x-axis at -4, -2.4916, -1, etc.

**Soroban** · October 2nd 2012, 07:30 PM

Hello, Jeka!

Are you familiar with the Extended Product Rule?

For three functions: . $y \;=\;f(x)\!\cdot\!g(x)\!\cdot\!h(x)$

. . we have: . $y' \;=\;{\color{blue}f'(x)}\!\cdot\!g(x)\!\cdot\!h(x) + f(x)\!\cdot\!{\color{blue}g'(x)}\!\cdot\!h(x) + f(x)\!\cdot\!g(x)\!\cdot\!{\color{blue}h'(x)}$

Do you see the pattern?

I need help find all the turning points of this function:

. . $f(x) \:=\: (x+1)^2(x-5)^3(x+4)^4(x-3)^5$

The derivative is: . $f'(x) \;=\;\begin{Bmatrix}{\color{blue}2(x+1)}\cdot(x-5)^3(x+4)^4(x-3)^5 \\ + (x+1)^2\cdot {\color{blue}3(x-5)^2}\cdot(x+4)^4(x-3)^5 \\ +(x+1)^2(x-5)^3\cdot {\color{blue}4(x+4)^3}\cdot (x-3)^5 \\ +(x+1)^2(x-5)^3(x+4)^4\cdot {\color{blue}5(x-3)^4} \end{Bmatrix} \;=\;0$

Factor: . $f'(x) \;=\;(x+1)(x-5)^2(x+4)^3(x-3)^4\cdot\begin{Bmatrix}2(x-5)(x+4)(x-3) \\ + 3(x+1)(x+4)(x-3) \\ + 4(x+1)(x-5)(x-3) \\ + 5(x+1)(x-5)(x+4) \end{Bmatrix} \;=\;0$

Some turning points are at: . $x \:=\:\text{-}1,5,\text{-}4,3$

I found no zeros for the cubic polynomial in the braces.
(But I see that MaxJasper found them.)

**Jeka** · October 2nd 2012, 07:52 PM

I don't really get it...
What i know right now is that, there are turning points at
(-1,0) & (5,0) & (-4,0) & (3,0)

There should be 3 or 4 more left
But i can't seem to find it.
And i'm talking about this function:
f(x) = (x+1)^(2)(x-5)^(3)(x+4)^(4)(x-3)^(5)
the above function, there are 4 four graphs (two types), combined together.
I want to know how i can find the other 3 or 4 turning points.

**MarkFL** · October 3rd 2012, 12:08 AM

You should ultimately be able to find:

$f'(x)=2(x+1)(x-5)^2(x+4)^3(x-3)^4(7x^3-15x^2-72x+22)$

Obviously, the roots from the linear factors are easy enough to find, but the roots of the cubic factor are not, and I would simply use Newton's method to approximate these roots. Let:

$g(x)\equiv7x^3-15x^2-72x+22$

Thus: $g'(x)=21x^2-30x-72$

So, we use the recursion:

$x_{n+1}=x_n-\frac{7x_n^3-15x_n^2-72x_n+22}{21x_n^2-30x_n-72}$

Looking at a graph of the cubic, we can then find reasonable guesses for the roots.

Root 1: $x_0\approx-2$

$x_1\approx-2.69\bar{4}$

$x_2\approx-2.50948582734$

$x_3\approx-2.49174430831$

$x_4\approx-2.49158454787$

$x_5\approx-2.49158453496$

$x_6\approx-2.49158453496$

Root 2: $x_0\approx0$

$x_1\approx0.30\bar5$

$x_2\approx0.290395504076$

$x_3\approx0.290370205766$

$x_4\approx0.290370205694$

$x_5\approx0.290370205694$

Root 3: $x_0\approx4$

$x_1\approx4.402\bar{7}$

$x_2\approx4.34537960674$

$x_3\approx4.34407214407$

$x_4\approx4.34407147212$

$x_5\approx4.34407147212$

So, we find the 3 roots of the cubic factor may be approximated by:

$x\approx-2.49158453496,0.290370205694,4.34407147212$

**Jeka** · October 3rd 2012, 05:17 PM

Sorry Guys, i seriously don't know where you guys are heading.
I don't know what rule you're using, i don't know how you guys got that.
It would be really really really helpful if you guys have some good explaination.

Soroban, i'm really happy that you're helping me.
But i don't know how you factored the first derivation of the function

MarkFL2, you too, i appreciate your help.
But i don't really know what you're doing. I don't know what rule. I know norhing.
You're just like, write so much stuff. That looks so pro. But i don't know what its say

Sorry, guys. Could you please have some explanation? Please!

Thanks

**MarkFL** · October 3rd 2012, 07:22 PM

Maybe things have changed since I was a student back in the days of stone-washed jeans and mullets, but we were taught Newton's method in Chapter 2 of Calc I.

**Jeka** · October 3rd 2012, 08:45 PM

MarkFL2, well you can teach me the way you were taught. I just want it solved

**MarkFL** · October 3rd 2012, 08:54 PM

If you read the link I provided, it will tell you how Newton's method works to converge to the root of a function.

If you just want it solved, then plug the values I gave for the roots into the original function to find the approximate values of the turning points resulting from the cubic factor of the derivative. If you wish for exact values, then read this article on how to obtain them:

Cubic function - Wikipedia, the free encyclopedia

**Jeka** · October 3rd 2012, 09:08 PM

But i'm not actually trying to find the x-intercepts on the x-axis. I'm trying to find the turning points. I already know all the x and y intercepts. What I am trying to find is the turning points, that aren't touching the x or y axis.

**MarkFL** · October 3rd 2012, 09:13 PM

The roots of the derivative give you the critical values. Roots of odd multiplicity will give you turning points, when plugged into the original function.

**Jeka** · October 3rd 2012, 09:31 PM

Sorry, I don't get it. Sorry.

Can you PLEASE do an example, on my function.
PLEASEEEEEEEEEEEEEEEEEEEEE!
I beg you!
Please!!!!!!!!

**MarkFL** · October 3rd 2012, 11:00 PM

I suggest working with simpler functions to get a feel for this first.

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