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
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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
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?
So you have the slope of the curve:
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.
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?
can anyone help me out here? did i do this right?
You have received two sggestions. You appear to have ignored both. Answer my previous question.
Find the second derivative of your original equation. Then, find the critical points. Finally, create a sign chart to see what the maximum is.Quote:
Originally Posted by Eric08
http://www.mathhelpforum.com/math-he...84-slopes.html <-----Check out this thread.
This is a take home test isn't it? Notice the "banned" label beneath Shinjiro's name? Be careful what you ask us.
take home test? Not trying to be mean but no its not, its just a question out of my book, and who is shinjiro?
-thanks for the hint