Turning points

**mathprob** · April 7th 2009, 02:25 PM

I dont know if this goes in here but this is a question from my pre-calc class that i could not solve

Find a polynomial whose turning points are at -1, (3 + √7)/4 , and (3 - √7)/4

Any help will be appreciated.

**Prove It** · April 7th 2009, 05:04 PM

Originally Posted by mathprob

I dont know if this goes in here but this is a question from my pre-calc class that i could not solve

Find a polynomial whose turning points are at -1, (3 + √7)/4 , and (3 - √7)/4

Any help will be appreciated.

Turning Points occur when the derivative is 0.

Using this information, you can find the derivative.

$\frac{dy}{dx} = (x + 1)\left[x - \left(\frac{3 + \sqrt{7}}{4}\right)\right]\left[x - \left(\frac{3 - \sqrt{7}}{4}\right)\right]$

(set it equal to 0, you should see that the turning points are correct).

Expand, and then you can integrate to find the polynomial.

$\frac{dy}{dx} = x^3 - \frac{1}{2}x^2 - x + \frac{1}{2}$

$y = \int{x^3 - \frac{1}{2}x^2 - x + \frac{1}{2}\,dx}$

$y = \frac{1}{4}x^4 - \frac{1}{6}x^3 - \frac{1}{2}x^2 + \frac{1}{2}x + C$ .

You can choose any value of C you like... 0 is the easiest.

So a polynomial that has the turning points you mentioned is

$y = \frac{1}{4}x^4 - \frac{1}{6}x^3 - \frac{1}{2}x^2 + \frac{1}{2}x$ .

**mathprob** · April 7th 2009, 05:35 PM

Originally Posted by Prove It

Turning Points occur when the derivative is 0.

Using this information, you can find the derivative.

$\frac{dy}{dx} = (x + 1)\left[x - \left(\frac{3 + \sqrt{7}}{4}\right)\right]\left[x - \left(\frac{3 - \sqrt{7}}{4}\right)\right]$

(set it equal to 0, you should see that the turning points are correct).

Expand, and then you can integrate to find the polynomial.

$\frac{dy}{dx} = x^3 - \frac{1}{2}x^2 - x + \frac{1}{2}$

$y = \int{x^3 - \frac{1}{2}x^2 - x + \frac{1}{2}\,dx}$

$y = \frac{1}{4}x^4 - \frac{1}{6}x^3 - \frac{1}{2}x^2 + \frac{1}{2}x + C$ .

You can choose any value of C you like... 0 is the easiest.

So a polynomial that has the turning points you mentioned is

$y = \frac{1}{4}x^4 - \frac{1}{6}x^3 - \frac{1}{2}x^2 + \frac{1}{2}x$ .

when i multiply
$\frac{dy}{dx} = (x + 1)\left[x - \left(\frac{3 + \sqrt{7}}{4}\right)\right]\left[x - \left(\frac{3 - \sqrt{7}}{4}\right)\right]$

i get 8x^3 - 4x^2 - 11x + 1

**mr fantastic** · April 7th 2009, 05:44 PM

Originally Posted by mathprob

when i multiply
$\frac{dy}{dx} = (x + 1)\left[x - \left(\frac{3 + \sqrt{7}}{4}\right)\right]\left[x - \left(\frac{3 - \sqrt{7}}{4}\right)\right]$

i get 8x^3 - 4x^2 - 11x + 1

After multiplying by 8, of course .....

Yes, I get the same result. Prove It made some small mistakes. But they don't affect the method he has shown you and you should be able to get the final answer without trouble.

**mathprob** · April 7th 2009, 05:56 PM

Originally Posted by mr fantastic

After multiplying by 8, of course .....

Yes, I get the same result. Prove It made some small mistakes. But they don't affect the method he has shown you and you should be able to get the final answer without trouble.

well, we never learned integrals thats why i could not use his method even though i know integrals

i got 12x^4 - 8x^3 - 33x^2 + 6x = 0

i need confirmation though, ill get some needed points for this

**mr fantastic** · April 7th 2009, 06:05 PM

Originally Posted by mathprob

well, we never learned integrals thats why i could not use his method even though i know integrals

i got 12x^4 - 8x^3 - 33x^2 + 6x = 0

i need confirmation though, ill get some needed points for this

Why are you doing questions that require integration if you never learned it?

Your answer is wrong. Read this: Integrals of Polynomials

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