# critical points

• Oct 26th 2009, 08:07 PM
burton6
critical points
f(x) = 18x + 15x^2 -4x^3 on the interval [-3,4]

A) f(-3) =
f(4) =

B) find the critical points of f(x) state x and y show your work

C) State the absolute maximum and absolute minimum values of the function on the interval [-3,4]

how do i get the critical points iv found the derivative to be 18+30x-12x^2 now what do i do?
• Oct 26th 2009, 08:15 PM
Mush
Quote:

Originally Posted by burton6
f(x) = 18x + 15x^2 -4x^3 on the interval [-3,4]

A) f(-3) =
f(4) =

B) find the critical points of f(x) state x and y show your work

C) State the absolute maximum and absolute minimum values of the function on the interval [-3,4]

You can't just post the question.

Where are you having trouble with it? Have you made an attempt? Where is your working?
• Oct 26th 2009, 08:23 PM
burton6
ow do i get the critical points iv found the derivative to be 18+30x-12x^2 now what do i do?
• Oct 26th 2009, 08:26 PM
Mush
Quote:

Originally Posted by burton6
ow do i get the critical points iv found the derivative to be 18+30x-12x^2 now what do i do?

Critical points are points at which the gradient of the function is zero. The gradient of a function f(x) is described by its derivative f'(x), so you're looking for the points at which f'(x) = 0.

So set f'(x) = 0, and solve for x. Once you have the values of x that solve this equation, plug them back into the original equation, f(x), to find the corresponding y values.
• Oct 26th 2009, 08:26 PM
Open that Hampster!
$f(x) = 18x + 15x^2 - 4x^3$
$f'(x) = 18 + 30x - 12x^2$

Critical numbers are numbers in an interval where f'(c) = 0

So just set the derivative to 0 and solve for the roots. If you can't evenly factor the equation then use the quadratic formula.

It'll make it easier on you if you factor 6 out of f'.
• Oct 26th 2009, 08:41 PM
burton6
tanks got it