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 decreasing?
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 a graph is the point where it intersects the y-axis. In this case, the y-axis is the line x = 0. Therefore, to find the y-intercept, we substitute x = 0 into the equation: f(0) = 4cos(0) = 4(1) = 4. So the y-intercept of f is 4.
b. To find where the graph of f is decreasing, we need to find the values of x where the derivative of f is negative. The derivative of f(x) = 4cos(x) with respect to x is f'(x) = -4sin(x). For -π ≤ x ≤ π, sin(x) is negative in the third and fourth quadrants. Therefore, -π/2 ≤ x ≤ 0 and π ≤ x ≤ 3π/2 are the values of x for which f is decreasing.
c. The absolute minimum value of f occurs when the cosine function achieves its minimum value, which is -1. Since f(x) = 4cos(x), the absolute minimum value of f is 4(-1) = -4.
d. To find the numbers x where f(x) = 0, we set 4cos(x) = 0 and solve for x. Dividing both sides by 4, we get cos(x) = 0. The cosine function equals 0 at π/2 and 3π/2. So, for 0 ≤ x ≤ 2π, the values of x that satisfy f(x) = 0 are π/2 and 3π/2.
e. The x-intercepts of f are the values of x where f(x) = 0. From the previous answer, we found that f(x) = 0 at x = π/2 and x = 3π/2. Therefore, the x-intercepts of f are π/2 and 3π/2.