Thread: Global max and Minimum

I have a few of these I'm practicing out of my book. Can you explain this to me a little better then the book. Here is their practice question:

f(x)=x^3+3x^2+1; -3<=x<=2


I would like to be able to solve their practice problems before pop quiz on Monday.

Hi

What you typically do in these cases are:

You differentiate the function and find max and minima.
But here you also need to test the endpoints of the interval.

You know how to perform a second derivative test to test for minima or maxima?

Some functions may not have a global maxima, if say a function is defined on , that is, the endpoints are not included, and the function in strictly growing, there might now exist a global maxima. Because whenever you pick a number close to 2, say 1.999 , you can always choose insted 1.999999 , and the function attains a higher value.

So keep a distinction between closed and open intervals.

so for t(x)=x^3+3x^2+1 -3<=x<=2

I found the critical points x=0 and x=-2

I plugged -3,-2,0,2 into the original function and came up with

absolute max (2,21)
absolute min (-3,-53)

Originally Posted by lisa1984wilson

so for t(x)=x^3+3x^2+1 -3<=x<=2

I found the critical points x=0 and x=-2

I plugged -3,-2,0,2 into the original function and came up with

absolute max (2,21)
absolute min (-3,-53)

as I can see f(0)=f(-3) right
so (0,1) and (-3,1) is local min

absolute max (2,21) it is correct as I see