Question

Absolute extrema of f(x) = tan(x) [3π/4, 5π/4]

a) Local minimum at x = 3π/4, local maximum at x = 5π/4
b) Local maximum at x = 3π/4, local minimum at x = 5π/4
c) Global minimum at x = 3π/4, global maximum at x = 5π/4
d) Global maximum at x = 3π/4, global minimum at x = 5π/4

Answer 1

The function f(x) = tan(x) on the interval [3π/4, 5π/4] has a global maximum at x = 5π/4 and a global minimum at x = 3π/4, considering the discontinuity at x = π.

Step-by-step explanation:

To determine the absolute extrema of the function f(x) = tan(x) on the interval [3π/4, 5π/4], we need to evaluate the function at the end points of the interval and also consider the continuity of the function within the interval.

First, we must recognize that the function tan(x) has a discontinuity (vertical asymptote) at x = π, which lies within the interval [3π/4, 5π/4]. Therefore, the function does not have a local minimum or maximum at the discontinuity, but we should check the end points for absolute extrema.

At x = 3π/4, f(x) = tan(3π/4) = -1, and at x = 5π/4, f(x) = tan(5π/4) = 1. Since the tangent function decreases on the interval from [3π/4, π) and increases on the interval from (π, 5π/4], the function has a global maximum at x = 5π/4 and a global minimum at x = 3π/4.