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. To find the y-intercept of the graph of f, we substitute x = 0 into the equation:
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.
f'(x) = -6 sin x
The graph of f is increasing when -π ≤ x ≤ π if -6 sin x > 0.
sin x is positive in the interval (-π, π), so -6 sin x > 0 for all x in the interval (-π, π).
Therefore, the graph of f is increasing for all numbers -π ≤ x ≤ π.
c. The absolute minimum value of f occurs when cos x = -1, which is -6.
Therefore, the absolute minimum value of f is -6.
d. To find when f(x) = 0, we solve the equation 6 cos x = 0 for x in the interval 0 ≤ x ≤ 2π.
cos x = 0 when x = π/2 and x = 3π/2.
So, in the interval 0 ≤ x ≤ 2π, f(x) = 0 at x = π/2 and x = 3π/2.
e. The x-intercepts of f occur when f(x) = 0.
We found in part d that f(x) = 0 at x = π/2 and x = 3π/2.
Therefore, the x-intercepts of f are x = π/2 and x = 3π/2.