# slope of a curve

• April 10th 2008, 12:23 PM
Eric08
slope of a curve
I need some help with this problem

What is the maximum slope of the curve y=4x^2-3x^3

- I got as far as y=8x-9x^2

now Im stuck, any help is appreciated
• April 10th 2008, 12:26 PM
TKHunny
The problem has fooled you.

What if you had been given this problem statement?

What is the maximum VALUE of f(x), given f(x) = 8x-9x^2

Could you solve that?
• April 10th 2008, 12:27 PM
o_O
So you have the slope of the curve: $y' = 4x^{2} - 3x^{3}$

Good. Now, if you want to find the maximum value of y' (the slope of the curve), what would you have to do?

It is the same process as what you would do if you were asked to find the maximum value that y has.
• April 10th 2008, 12:33 PM
Eric08
well i plugged it into my calculator and found the maximum value for the parabola. It came out to be x=.444 and y=1.77. Is this all they are asking for? how would i go about doing it without the calculator?
• April 10th 2008, 01:31 PM
Eric08
can anyone help me out here? did i do this right?
• April 10th 2008, 01:39 PM
TKHunny
You have received two sggestions. You appear to have ignored both. Answer my previous question.
• April 10th 2008, 01:39 PM
topher0805
Quote:

Originally Posted by Eric08
can anyone help me out here? did i do this right?

Find the second derivative of your original equation. Then, find the critical points. Finally, create a sign chart to see what the maximum is.
• April 10th 2008, 01:41 PM
topher0805