# points where function has a horizontal tangent line

• Jan 11th 2009, 09:51 PM
Neverending
points where function has a horizontal tangent line
Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.

http://i146.photobucket.com/albums/r...kitty216/3.jpg

First I plugged in o to all the x's and got the point (2,0).. I don't know how to find the other two points. I want to learn the method of how to work these types of problems out, then I can do the same types of problemsusing the same method. Help would be great!
• Jan 11th 2009, 10:00 PM
mr fantastic
Quote:

Originally Posted by Neverending
Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.

http://i146.photobucket.com/albums/r...kitty216/3.jpg

First I plugged in o to all the x's and got the point (2,0).. I don't know how to find the other two points. I want to learn the method of how to work these types of problems out, then I can do the same types of problemsusing the same method. Help would be great!

Horizontal tangent means gradient of tangent is equal to zero.

Solve dy/dx = 0 to get the x-coordinates of the required points.
• Jan 11th 2009, 10:08 PM
Neverending
dy/dx means find derivative of..

so how do I find where the derivative equals 0?
• Jan 11th 2009, 10:13 PM
mr fantastic
Quote:

Originally Posted by Neverending
dy/dx means find derivative of..

so how do I find where the derivative equals 0?

You start by calculating what the derivative is. Then you set it equal to zero. Try posting what you can do and where you get stuck and it will be easier to help you.
• Jan 11th 2009, 10:13 PM
Prove It
Quote:

Originally Posted by Neverending
Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.

http://i146.photobucket.com/albums/r...kitty216/3.jpg

First I plugged in o to all the x's and got the point (2,0).. I don't know how to find the other two points. I want to learn the method of how to work these types of problems out, then I can do the same types of problemsusing the same method. Help would be great!

If $\displaystyle y = ax^n$ then $\displaystyle \frac{dy}{dx} = nax^{n-1}$.

Using this, we get

$\displaystyle \frac{dy}{dx} = 4x^3 - 16x$.

We want to find the x-values where this derivative is 0. So we LET $\displaystyle \frac{dy}{dx} = 0$ and solve for x.

$\displaystyle 0 = 4x^3 - 16x$

$\displaystyle 0 = 4x(x^2 - 4)$

$\displaystyle 0 = 4x(x + 2)(x - 2)$

$\displaystyle 0 = 4x$ or $\displaystyle 0 = x + 2$ or $\displaystyle 0 = x - 2$.

So $\displaystyle x = 0$ or $\displaystyle x = -2$ or $\displaystyle x = 2$ are the points we want.
• Jan 11th 2009, 10:19 PM
Neverending
oh so I find the derivative first.. i see.. thanks guys!