1. Derivatives

I'm supposed to find the maximum slope of the curve $y = 8x^3 + 2x^2$

My work:

$y' = 24x^2 + 4x$

$y'' = 48x + 4$

$24x^2 + 4x = 0$

$4x(6x + 1) = 0$

$6x + 1 = 0$

$x = \frac{-1}{6}$

My question is, am I supposed to solve for x in $y^{''}$ as well or no? Supposing I'm not but do I plug the value I got from x back into the original equation to solve for y then that gives me the maximum slope of the curve?

2. I'm supposed to find the maximum slope of the curve $y = 8x^3 + 2x^2$

My work:

$y' = 24x^2 + 4x$

$y'' = 48x + 4$
This is all correct. Your next step should be finding when the second derivative is equal to 0:

$48x + 4=0$

$48x=-4$

$x=-\frac {1}{12}$

Now plug that back into the first derivative:

$
f'(-\frac {1}{12})=24(-\frac {1}{12})^2 + 4(-\frac {1}{12})$

3. If you solved for y' = 0, you found the local maximum (or minimum, whichever it may be) for y. If you want to find the maximum slope, i.e. when y' is maximum, how would you use y'' to help you find that?

4. Oh so its just solve for x in second derivative then plug that number in the first derivative and that gives me the maximum slope of the original? If so cool

Edit: When I plugged $\frac{-1}{12}$ into the first derivative I got -.166. What does this tell me exactly. The maximum slope, or minimum?

5. Originally Posted by R3ap3r
Oh so its just solve for x in second derivative then plug that number in the first derivative and that gives me the maximum slope of the original? If so cool
Yes. Think about the definition of the derivative. It is the slope of the original curve. So when we are looking for the max of the slope of the curve, we are simply looking for the max of the derivative. We can use the same method to find this that we use to find the max and mins of a curve.

6. Do you think I should write my answer as the Maximum slope of the curve is $-.166$ or $y' \left(\frac{-1}{12}\right) = -.166$

7. Originally Posted by R3ap3r
Do you think I should write my answer as the Maximum slope of the curve is $-.166$ or $y' \left(\frac{-1}{12}\right) = -.166$
$y' \left(\frac{-1}{12}\right) = -.166$

More mathy.

If that's a word...

8. more mathmetical? perhaps