Let f(x) = 4cosx
Complete parts (a) through (e) below
a. What is the y-intercept of the graph of f?
b. For what numbers x, -π ≤ x ≤ π, is the graph of f increasing?
c. What is the absolute maximum value of f?
d. For what numbers x, 0 ≤ x ≤ 2π, does f(x) = 0?
e. What are the x-intercepts of f?
a. To find the y-intercept, we set x = 0 and evaluate f(0):
f(0) = 4cos(0) = 4(1) = 4
So the y-intercept of the graph of f is 4.
b. The graph of f is increasing when the derivative f'(x) is positive. We find the derivative of f(x) using the chain rule:
f'(x) = -4sin(x)
Since sin(x) is positive when x is between -π/2 and π/2, f'(x) = -4sin(x) is positive when x is between -π/2 and π/2. Therefore, the graph of f is increasing when -π/2 < x < π/2.
c. The absolute maximum value of f occurs when cos(x) = 1, which happens when x = 0. So the absolute maximum value of f is f(0) = 4.
d. To find when f(x) = 0, we set 4cos(x) = 0 and solve for x. Using the unit circle, we see that cosine is 0 at x = π/2 and x = 3π/2. However, we are only concerned with the range 0 ≤ x ≤ 2π, so the solutions are x = π/2 and x = 3π/2.
e. The x-intercepts of f occur when f(x) = 0. Using the solutions from part (d), the x-intercepts are x = π/2 and x = 3π/2.