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