find local max and min

Question

f(x)= cos^2(x-pi/4) on the interval (0, 2pi)

Answer 1

We have to set the derivative of the function equal to zero.  This is because the slope of the line tangent to the maximum and minimum is zero.

 

d/dt[cos²(x - π/4

 

-2cos(x - π/4)sin(x - π/4) = 0

 

-2[(cos(x)cos(π/4) + sin(x)sin(π/4))(sin(x)cos(π/4) - cos(x)sin(π/4))] = 0

 

 

We will set the following equal to zero:

 

cos(x) = 0                       sin(x) = 0                

   

x = π/2                            x = 0              

x = 3π/2                          x = π

                                       x = 2π

 

Our critical points are

x = 0, π/2, 3π/2, 2π

 

 

Now we perform a test point:

 

We substitute the values of x greater than or less than each critical point when we evaluate the derivative of the function.  This will allow us to determine where on the graph there is an increase or decrease.  For instance, if the value of the derivative is negative, then the graph is decreasing.  Decrease to increase indicates a minimum, and increase to decrease indicates a maximum.

 

Then substitute the values of the critical points into the original function to find the y value of the points.  These will be your minimum and maximum values.

 

Answer 2

Raj, not so bad.

First, think of what the cos(x) function looks like: it's 1 at x=0, curves off down to 0 at x=pi/2, reaches -1 at x=pi, and runs back up through 0 at x=3pi/4, and on up to 1 again at x=2pi.

All right, now look at what the argument is: (x-pi/4) when x=0: -pi/4 puts you at -45.degrees., where cos is sqrt(2)/2, incidentally -- but over any 2pi wide interval, you're still running the cos(x-pi/4) all the way through the range -1 to 1, inclusive -- it looks like the cosine curve, but shifted right by pi/4. So the cos^2 is going to run from 0 to 1, inclusive, in the same interval. I guess that's a local max of 1 at x=pi/4 and 3pi/4, and local min of 0 at 3pi/4, 7pi/4. You don't even need to fret about the sloppy use of (0, 2pi) as domain (sloppy, because usually one of the endpoints would be included, for example, as (0, 2pi] -- that way, if the function maximizes there, you won't be left wondering what's the maximum if the actual endpoint is excluded!).