1. ## 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.

2. 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

3. 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

4. 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

5. 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

6. ## Re:

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:

7. 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