1. ## Concavity

Hi All,

I have a concavity question and my work. I'm having some trouble finding the absolute maximum.

Question:

Code:
let f(x)=x^3-(3/2)x^2 on the interval [-1,2].
Find the absolute maximum and absolute minimum
of f(x) on this interval.
I took the derivative and found the absolute minimum at -1. Are you supposed to take the second derivative to get the absolute max? I also tried to factor the derivative but was unsuccessful.

anyone remember doing this?

-M

2. Originally Posted by mant1s
Hi All,

I have a concavity question and my work. I'm having some trouble finding the absolute maximum.

Question:

Code:
let f(x)=x^3-(3/2)x^2 on the interval [-1,2].
Find the absolute maximum and absolute minimum
of f(x) on this interval.
I took the derivative and found the absolute minimum at -1. Are you supposed to take the second derivative to get the absolute max? I also tried to factor the derivative but was unsuccessful.

anyone remember doing this?

-M
$\frac{d}{dx}(x^3-\frac{3}{2}x^2)=3x^2-3x$

Setting the derivative to zero will give all relative maxima and minima for the function

$3x(x-1)=0\Rightarrow{x=1,0}$

bUT THIS MAY NOT HAVE ANY BEARING ON OUR PROBLEM. These points are both in the interval, but the endpoints could very well be what we are after. to find them

$\lim_{x\to{-1^+}}f(x)=-1-\frac{3}{2}=-\frac{5}{2}$

and

$\lim_{x\to2^-}f(x)=8-6=2$

And you can see that the absolute max and min occur at the endpoints of the interval.

The key here is to know what the problem is asking, and then know what you need to use to solve it.

3. $f'(x)=3x^2-3x=3x(x-1)$

put $f'(x)=0$ to get $x=0,1$

maximum value will be maximum of $f(-1),f(0),f(1),f(2)$

and the minimum value will be minimum of $f(-1),f(0),f(1),f(2)$