f(x) = cos²1.45x on 0

Question

f(x) = cos²1.45x on 0 < x < 2 $\pi$ has A) a relative maximum only B) a relative minimum only C) both a relative maximum and a relative minimum D) neither a relative maximum nor a relative minimum E) cannot be determined

Quizplus · Accepted Answer

f(x) is a composition of trigonometric functions, which are continuous on the given interval. Therefore, f(x) is also continuous. Taking the derivative of f(x), we get: f'(x) = -sin(²1.45x) * ²1.45 Setting f'(x) = 0 to find critical points, we get: -sin(²1.45x) * ²1.45 = 0 Since sin(²1.45x) can never be 0, the critical points are where ²1.45x = F1F1F1S1F1F1F1/2 or x = F1F1F1S1F1F1F1/2².723. At x = F1F1F1S1F1F1F1/2².723, f''(x) = -²1.45sin(F1F1F1S1F1F1F1/2².723), which is negative, indicating a relative maximum. At x = 0 and x = F1F1F1S1F1F1F1/2, f''(x) = -²1.45sin(0) and ²1.45sin(S1), respectively, which are both equal to 0. This indicates that there are horizontal inflection points at x = 0 and x = F1F1F1S1F1F1F1/2. At x = F1F1F1S1F1F1F1, f''(x) = -²1.45sin(S1), which is also equal to 0. However, since f''(x) changes sign from negative to positive at this point, it is not a relative extremum. Therefore, f(x) has both a relative maximum and a relative minimum.

F(x) = Cos21 π\piπ Has A) a Relative Maximum Only B) a Relative Minimum

F(x) = Cos²1 $\pi$ Has
A) a Relative Maximum Only
B) a Relative Minimum