On-line Math 21

On-line Math 21

4.1  The Mean Value Theorem

Example 1 Find the local maximum and minimum points and values, and the absolute maximum minimum points and values, of the function
f(x) = 2sin(x)+cos(2x), x Î [0,2p].

Solution

To find the critical points, we first differentiate the function,
f¢(x) = 2cos(x)-2sin(2x),
and we solve the equation f¢(x) = 0 ,
0
=
f¢(x)
=
2cos(x)-2sin(2x)
=
2cos(x)-4sin(x)cos(x) since  sin(2x) = 2sin(x)cos(x)
=
2cos(x)(1-2sin(x)),
so that
f¢(x) = 0Û cos(x) = 0 or sin(x) = 1/2.
Solving for x at this stage is simply a matter of remembering where cos(x) = 0 (which is x = p/2, x = 3p/2 only, in the domain [0,2p] ), or where sin(x) = 1/2 ( x = p/6 or x = 5p/6 only, in the interval).

According to the theorem mentioned earlier, called Fermat's Theorem, if the derivative exists at a maximum or minimum point (local or absolute), then the derivative has to be 0. The only other options are places where the derivative does not exist, or the endpoints of the interval. In this case, though, the derivative exists at all x , so the only possibilities for maximum or minimum points are at these places where the derivative is 0, and at the endpoints, 0 and 2p.

So, how do you tell which is a max or a min? First off, evaluate the function at all of these points. Since:
f(x)
=
2sin(x)+cos(2x)
f(0)
=
2·0+1 = 1
f(p/6)
=
2· 1
2
 + 1
2
 = 3
2
f(p/2)
=
2·1+(-1) = 1
f(5p/6)
=
2· 1
2
 + 1
2
 = 3
2
f(3p/2)
=
2·(-1)+(-1) = -3
f(2p)
=
2·0+1 = 1,
the absolute maximum value is 3/2 , achieved at x = p/6 and x = 5p/6 . The absolute minimum value is -3 , achieved at x = 3p/2 .

What about the points in between? Which are local maxima, and which are local minima? Well, since the function has to increase up to, and decrease away from, a local maximum,

then f¢(x) > 0 for x < x0 and f¢(x) < 0 for x > x0 at a local maximum point x0 , as long as the x doesn't get too far from x0 . Similarly, f¢(x) < 0 for x < x0 and f¢(x) > 0 for x > x0 at a local minimum point x0 , as long as the x doesn't get too far from x0 . In this case, we can look at the sign of the first derivative to see what happens. Since
f¢(x) = 2cos(x)(1-2sin(x)),
you can see that f¢(x) will have the following signs:

which clearly shows that x = p/6 is a local maximum (not surprizing, since it is an absolute maximum), x = p/2 is a local minimum, x = 5p/6 is a local max, and x = 3p/2 is a local min. So, that left-over point, x = p/2 , is a local minimum point. We don't usually call an endpoint a ``local'' maximum or minimum, but an endpoint might be an absolute maximum or minimum.

Copyright (c) 2000 by David L. Johnson.

File translated from TEX by TTH, version 2.61.
On 5 Dec 2000, 01:11.