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 maximum value of​ f?

d. For what numbers​ x, 0 ≤ x ≤ 2π​, does ​f(x)​ = 0?

e. What are the​ x-intercepts of​ f?

1 answer

a. The y-intercept of a graph is the value of the function when x = 0. Substituting x = 0 into f(x) = 4cosx, we get f(0) = 4cos(0) = 4(1) = 4. Therefore, the y-intercept of the graph of f is 4.

b. The graph of f = 4cosx is decreasing when the derivative of f with respect to x is negative. Taking the derivative of f(x) = 4cosx, we get f'(x) = -4sinx. The graph of f is decreasing when -π ≤ x ≤ π for x values in the interval (-π/2, π/2).

c. The cosine function has a maximum value of 1. Therefore, the absolute maximum value of f is 4(1) = 4.

d. For f(x) = 0, we need to solve the equation 4cosx = 0. Using the unit circle or the properties of the cosine function, we know that cos(x) = 0 when x = π/2 or x = 3π/2. Therefore, for 0 ≤ x ≤ 2π, f(x) = 0 when x = π/2 or x = 3π/2.

e. The x-intercepts of a graph are the values of x for which f(x) = 0. From part d, we found that f(x) = 0 when x = π/2 or x = 3π/2. Therefore, the x-intercepts of f are π/2 and 3π/2.