# Setting g'(x) to zero

• Mar 21st 2008, 02:42 PM
Boris B
Setting g'(x) to zero
(I know this is the pre-cal forum, but the calculus isn't my problem here ... it's the basics that are dogging me.)
In order to make a wiggle graph, I need to find where my slopes are zero. The formula of my derivative is:
$g'(x) = 6x^2 - 7x - 3$

So my question is, when g'(x) = 0, x = ?

I don't think I'm just supposed to plug random numbers in and hope to hit 0, but simplifying has only gotten me a little closer. I think I can simplify to:

$x = 1/(2x) + 7/6$

It would be a little easier to just start plugging numbers in, but I think I'm supposed to actually solve this one before moving on. I'm stuck - how do I know what x is when g'(x) = 0?
• Mar 21st 2008, 02:48 PM
mr fantastic
Quote:

Originally Posted by Boris B
(I know this is the pre-cal forum, but the calculus isn't my problem here ... it's the basics that are dogging me.)
In order to make a wiggle graph, I need to find where my slopes are zero. The formula of my derivative is:
$g'(x) = 6x^2 - 7x - 3$

So my question is, when g'(x) = 0, x = ?

I don't think I'm just supposed to plug random numbers in and hope to hit 0, but simplifying has only gotten me a little closer. I think I can simplify to:

$x = 1/(2x) + 7/6$

It would be a little easier to just start plugging numbers in, but I think I'm supposed to actually solve this one before moving on. I'm stuck - how do I know what x is when g'(x) = 0?

You need to solve the quadratic equation

$0 = 6x^2 - 7x - 3$ .....