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
