These observations lead to a generalization for any function f(x) that has a derivative on an interval I:

     Important stuff!! 

  1)  If f '(x) > 0 on an interval I, then the graph of f(x) rises as x increases.

  2)  If f '(x) < 0 on an interval I, then the graph of f(x) falls as x increases.

  3)  If f '(c) = 0, then the graph of f(x) has a horizontal tangent at x = c.  The function may have a local maximum or minimum value, or a point of inflection.

 

2)  To be a maximum point, the graph must change direction from increasing to decreasing.

3)  To be an inflection point, the graph doesn't change direction.  In the above example ( one in middle) it is increasing before the f '(c) = 0 and it is still increasing after.  You can also have one with the graph decreasing on both sides.

a)  Find the critical points by finding f '(x).

Find the zeros by solving f '(x) = 0

Therefore, we have critical points at x = 0, and x = -2

Substitute these values into the original function to find the y values of the critical points. The points are (0, -4) and (-2, 0)

b)  Use the derivative to find where the graph is increasing and decreasing by taking x values in each of the three areas formed by the two critical points.  The chart below shows the results
 

	f '(x)
x < -2	+
x = -2	0
-2 < x < 0	-
x = 0	0
x > 0	+

The chart shows that (-2, 0) is a local Max. and (0, -4) is a local min.  You can tell because of the sign changes!

c)  Find the zeros of the original function.  These are the x-intercepts.  You can use synthetic division and factoring to find the zeros!  They are (-2, 0) (double root) and (1, 0)

d)  Find the y-intercept.  This is the constant of the original function.  (0, -4)

e)  Now take the limit as x goes to both infinities of the original function.   

f)  Now put all these together and graph the function!

a)  Find the critical points by finding f '(x)

Set f '(x) = 0 and solve

Therefore, the critical points are at x = 0, x = 2 and x = -2
Substitute these into the original function and the points are (0, 9), (2, -7), (-2, -7)
b)  Use the derivative to find the slope in each of the four areas divided by the three critical points.  The chart is below  

	f '(x)
x < -2	-
x = -2	0
-2 < x < 0	+
x = 0	0
0 < x < 2	-
x = 2	0
x > 2	+

 
The chart shows that (-2, -7) and (2, 7) are minimums and (0, 9) is a maximum.  Check out the signs on the chart to realize this!
c)  Find the zeros of the original function by using your graphing calculator.  The zeros are not rational but irrational.

d)  The y intercept is (0, 9)

e)  Find the limits on infinity.   

f)  Put these together and graph the function!

      The next section is one you are going to love!  Yes, your favorite and mine.  Word problems related to maxima and minima!!  Can't wait!!