Find and classify the global extrema of the following function:
f(x) = 7 cos(x)
f(x) = 7 cos(x) has period 2 π
Look for extrema over the domain of f(x) restricted to the interval 0<=x<2 π:
f(x) = 7 cos(x) when 0<=x<2 π
Find the critical points of f(x):
Compute the critical points of 7 cos(x)
To find all critical points, first compute f'(x):
( d)/( dx)(7 cos(x)) = -7 sin(x):
f'(x) = -7 sin(x)
Solving -7 sin(x) = 0 yields x = 0 or x = π:
x = 0, x = π
f'(x) exists for all x such that 0<=x<2 π:
-7 sin(x) exists for all x such that 0<=x<2 π
The critical points of 7 cos(x) occur at x = 0 and x = π:
x = 0, x = π
The domain of 7 cos(x) on 0<=x<2 π is {x element R : 0<=x<2 π}:
The endpoints of {x element R : 0<=x<2 π} are x = 0 and 2 π
Evaluate 7 cos(x) at x = 0, π and 2 π:
The open endpoints of the domain are marked in gray
x | f(x)
0 | 7
π | -7
2 π^- | 7
The largest value corresponds to a global maximum, and the smallest value
corresponds to a global minimum:
The open endpoints of the domain are marked in gray
x | f(x) | extrema type
0 | 7 | global max
π | -7 | global min
2 π^- | 7 | global max
Remove the points x = 2 π^- from the table
These cannot be global extrema, as the value of f(x) here is never achieved:
x | f(x) | extrema type
0 | 7 | global max
π | -7 | global min
Add 2 π k for k element Z to each extremum:
x = 0->x = 2 π k for k element Z
x = π->x = 2 π k + π for k element Z
f(x) = 7 cos(x) has a family of global minima and a family of global maxima:
Answer: | f(x) has a global maximum at x = 2 π k for k element Z
f(x) has a global minimum at x = 2 π k + π for k element Z
