Find the absolute extrema of the function on the closed interval.
f(x) = x^3 − (3/2)x^2 [−1, 6]

minimum (x,y) =
maximum (x,y) =

Question

Find the absolute extrema of the function on the closed interval.
f(x) = x^3 − (3/2)x^2     [−1, 6]

minimum (x,y) = 
maximum (x,y) =

mathmale · Answer

This problem has two parts.  One part is to determine any relative minima/maxima within the interval [-1,6].  The other is to determine the values of the given function at the endpoints.  Which of these are you able to do now?

anonymous · Answer

I differentiated the function as 3x^2-2(3/2)x
This became 3x(1-x)

mathmale · Answer

Agreed, except that I'd reverse the quantity within parentheses:
3*x*(x-1).  Can you agree with that?

anonymous · Answer

yes

mathmale · Answer

Ok.  that's your derivative.  set this = to 0 and solve for y our critical values.  What are they?

anonymous · Answer

3x(x-1)=0
x=0,1
f(0)= -1
f(1)=0
f(-1)=6
f(6)=90
critical values = -1,90

mathmale · Answer

Your critical values are actually {0,1}.  These are the roots of the equation f '(x)=0, right?
When i evaluated f(x) = x^3 − (3/2)x^2 at -1, I got -2.5; at 6, I got 162.
Would you mind double-checking your work there?

mathmale · Answer

Is this correct?$$f(x) = x^3 − (3/2)x^2 $$

anonymous · Answer

yes

mathmale · Answer

Then$$f(x) = (-1)^3 − (3/2)(-1)^2 =?$$

anonymous · Answer

2.25

mathmale · Answer

Here's what I get:  f(-1)=-1-(3/2)(1)=-2.5.

mathmale · Answer

We need to agree on all the y-values before we can proceed.

mathmale · Answer

$$f(6) = (6)^3 − (3/2)(6)^2 =?$$

anonymous · Answer

162

mathmale · Answer

I agree with that.
Let's now make a table:|dw:1395446935464:dw|