I don't understand....... The questions is: Use a graphing calculator to find the intervals on which the function f(x)=x^3-3x^2+2 is increasing or decreasing and find any maxima or minima. Can someone tell me how to do this??

So it looks like you have a max at (-1,4) and a min at (1,0).

Your intervals would be as follows:

f(x) is increasing when x<-1 or x>1 also written in set notation: (-infinity,-1) or (1, infinity)

f(x) is decreasing when: -1<x<1 also written in set notation (-1,1)

See below how to get this on TI-89. You can message me if you have a different type of calculator.

**TI-89**

1) Press [◊] [Y=] plug x^3-3x^2+2 as if you were about to graph the function

2) Graph

3) Go to F5 and then down to 4:Maximum

4) It will then prompt you to do an lower bound in which you will bring the cursor closest to the maximum point from the right by using the arrows on your calculator. Enter.

5) Then it will ask for upper bound, so do the same thing but from the left of the maximum. Enter.

The calculator should give you the maximum

For the minimum do the same thing but when you get to F5 go to minimum.

For increasing and decreasing, you can do this by just looking at the graph and looking at where it is increasing and decreasing. So as you can see the graph increases until you get to (-1,4). So the graph is increasing when x is less than -1. Write
this like -1<x. It is also increasing when x is greater than 1. So the graph is increasing when x is less than -1 and when x is greater than 1. This is written like this:

x is increasing when x>-1 OR x<1. The word OR is used here because it is a union of two sets. The sets are all the real numbers less than -1 and greater than 1.

Then for decreasing, notice that the graph starts decreasing at (-1,4) and then stops decreasing at (1,0). Again only use the x values to express this. So f(x) decreases when -1<x<1.