Let f(x) = 6 cos x.
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 minimum value of f?
d. For what numbers x, 0 ≤ x ≤ 2π, does f(x) = 0?
e. What are the x-intercepts of f?
a. The y-intercept of the graph of f is the value of f when x = 0. So, substituting x = 0 into f(x) = 6 cos x, we get f(0) = 6 cos 0 = 6(1) = 6. Therefore, the y-intercept is 6.
b. The graph of f is increasing when the derivative of f is positive. The derivative of f(x) = 6 cos x is f'(x) = -6 sin x. For the graph of f to increase, we need -6 sin x > 0. This inequality is satisfied when -π/2 < x < π/2. In other words, for -π < x < -π/2 and π/2 < x < π, the graph of f is increasing.
c. The absolute minimum value of f is the lowest possible y-value of the graph. Since the range of the cosine function is -1 to 1, the absolute minimum value occurs when cos x = -1. This happens when x = π. So, the absolute minimum value of f is f(π) = 6 cos π = 6(-1) = -6.
d. To find the values of x where f(x) = 0, we solve the equation 6 cos x = 0. In the range 0 ≤ x ≤ 2π, cos x = 0 when x = π/2 and x = 3π/2. So, the values of x where f(x) = 0 are x = π/2 and x = 3π/2.
e. The x-intercepts of f are the values of x where f(x) = 0. Therefore, the x-intercepts are x = π/2 and x = 3π/2.