# Derivatives Problem

• Oct 17th 2007, 07:43 AM
FalconPUNCH!
Derivatives Problem
For what values of x does the graph of:

F(X) = X^3 + 3X^2 + X + 3 have a horizontal tangent?

I don't know how to approach this but I got 0 as one of the values of X. I'm not really sure if there are other values.
• Oct 17th 2007, 07:52 AM
topsquark
Quote:

Originally Posted by FalconPUNCH!
For what values of x does the graph of:

F(X) = X^3 + 3X^2 + X + 3 have a horizontal tangent?

I don't know how to approach this but I got 0 as one of the values of X. I'm not really sure if there are other values.

The slope of the tangent is the derivative, so
$\displaystyle F^{\prime}(X) = 3X^2 + 6X + 1$

You are looking for a horizontal tangent, so you are looking for a slope of 0. Thus you are looking for a value of X such that
$\displaystyle 3X^2 + 6X + 1 = 0$

-Dan
• Oct 17th 2007, 06:33 PM
FalconPUNCH!
Quote:

Originally Posted by topsquark
The slope of the tangent is the derivative, so
$\displaystyle F^{\prime}(X) = 3X^2 + 6X + 1$

You are looking for a horizontal tangent, so you are looking for a slope of 0. Thus you are looking for a value of X such that
$\displaystyle 3X^2 + 6X + 1 = 0$

-Dan

So do I have to use the quadratic formula in order to do it? If so then I did it earlier today
• Oct 17th 2007, 06:36 PM
Jhevon
Quote:

Originally Posted by FalconPUNCH!
So do I have to use the quadratic formula in order to do it? If so then I did it earlier today

yes, that is one way. there's no nice way to factor this as there are no rational roots
• Oct 17th 2007, 06:39 PM
vesperka
If you have a graphing calculator like a TI-83 you can type in the equation for the derivative and then hit 2nd --> Calc to find the zeros. I'm guessing your teacher probably wants to see work and it doesn't hurt to practice by hand, but it's a good way to check your answers :)
• Oct 17th 2007, 07:04 PM
qbkr21
Re:
Quote:

Originally Posted by FalconPUNCH!
For what values of x does the graph of:

F(X) = X^3 + 3X^2 + X + 3 have a horizontal tangent?

I don't know how to approach this but I got 0 as one of the values of X. I'm not really sure if there are other values.

RE:
• Oct 18th 2007, 06:52 AM
FalconPUNCH!
Quote:

Originally Posted by Jhevon
yes, that is one way. there's no nice way to factor this as there are no rational roots

I just did it with the quadratic formula. Thank you guys for helping