Final Answer:
The critical numbers of f in [0,2π] are π/2 and 3π/2.
Step-by-step explanation:
To find the critical numbers of a function, we need to find the values of x where the derivative does not exist or is equal to zero. In this case, the function is f(θ) = 4θ - tanθ.
First, let's find the derivative:
f'(θ) = 4 - sec²θ
Now, we need to find the values of θ where f'(θ) is equal to zero or does not exist. The secant function is defined for all real numbers except for π/2 and 3π/2 (radians). Therefore, at these points, sec²θ becomes infinite, which means that f'(θ) does not exist.
When sec²θ = ±∞, then tanθ = ±1. This happens at π/2 and 3π/2 radians. Therefore, the critical numbers of f in [0,2π] are π/2 and 3π/2.
At these critical points, the behavior of the function changes drastically. When θ approaches either π/2 or 3π/2 from the left or right, f(θ) approaches positive infinity. However, when θ passes through either of these critical points, f(θ) jumps to negative infinity and continues decreasing as θ increases beyond these points.
This behavior is due to the fact that as tanθ approaches ±1 from either side of these critical points, sec²θ becomes very large in magnitude but alternates signs infinitely many times as tanθ approaches ±1 from either side. This causes f(θ) to oscillate between positive and negative infinity infinitely many times as we pass through these critical points.