Lori Pearman
EMT 725

Problem: Find the Maximum of f(x) = (1-x)(1+x)(1+x).

One approach to this problem is to graph the function on a graphing calculator or a computer graphing tool in order to obtain an estimate of the maximum. I used Algebra Xpresser to graph f(x).
This is part of the graph that I got (it shows the interval from 0 to 1):

From this graph, I can estimate that the maximum is about 1.18. This maximum occurs when x is about 0.33. One can keep zooming in on the "top of the hill" in the picture to get better estimates.

Calculus could be used to find the maximum of f'(x) more exactly.


Now we need to set f '(x) = 0 to get the critical values.

f '(x) = (3x-1)(x+1) = 0

There are possible relative maximum (or minimum)when x = 1/3 or x = -1.

We can now use the first or second derivative test to determine where we have a relative maximum.
Note: -1 is not in our interval from 0 to 1, but I will check it anyway.
f ''(x)=-6x-2
I will use the second derivative test.
f ''(-1) is a postive value. This implies that when x=-1, the graph of f (x) is concave upward.
Thus, f (-1) is a relative minimum.
f ''(1/3) is a negative value. This implies that when x=1/3, the graph is concave downward.
f (1/3) is a relative maximum.

In the interval from 0 to 1, there is a maximum which is f (1/3)=1.185185

Now let's use a spread sheet to estimate the maximum of f(x) in the interval from 0 to 1.
I could tell from my Algebra Xpresser graph that in the interval [0,1], the maximum occured somewhere between x=0.2 and x=0.5. Below is a spread sheet table with x-values in increments of 0.01 that are in this interval.

If I want an even better approximation, I can look at a smaller interval, such as [0.32,0.34] .
(I can tell from the previous spread sheet that this interval contains the maximum. ) Now that I've narrowed down the interval containing the maximum, it is easier to look at even smaller increments,
thus obtaining a more accurate approximation of the maximum value. (See below table.)

From this table, we can see that the maximum occurs approximately when x=0.333.

One could continue this process of "zooming in" on the maximum value by using smaller intervals and smaller increments for each new table. The more times this is done, the more accurate the maximum value will be.

The arithmetic mean- geometric mean inequality can also be used to find the maximum value.
f(x) = (1-x)(1+x)(1+x)
= [2(1-x)(1+x)(1+x)]/2
= (1/2)[(2-2x)(1+x)(1+x)]

For non-negative terms, the inequality gives:
(1/2)[(2-2x+1+x+1+x)/3]^3 > or = [(1/2)(2-2x)(1+x)(1+x)]

So (1/2)[4/3]^3 > or = f(x)
1.185185 > or = f(x) with equality iff (2-2x)= (1+x).
The maximum value of f(x) is about 1.185, and this occurs when x= 1/3.


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