Let ​f(x) = 4cosx

Question

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?

Answer 1

To find the y-intercept of the graph of f(x) = 4cos(x), we need to find the value of f(0). Plug in x = 0 into the equation:

f(0) = 4cos(0) = 4(1) = 4

So, the y-intercept of the graph of f is 4.

To find when the graph of f is decreasing, we need to find the interval on which the derivative of f is negative. Take the derivative of f(x) with respect to x:

f'(x) = -4sin(x)

Set f'(x) < 0:

-4sin(x) < 0

Since sin(x) is negative in the interval -π ≤ x ≤ 0 and π ≤ x ≤ 2π, the graph of f is decreasing in those intervals.

The absolute maximum value of f occurs at the highest point of the graph. Since the range of cosine function is -1 to 1, the absolute maximum value of f is 4.

To find where f(x) = 0, we need to solve the equation:

4cos(x) = 0

cos(x) = 0

The solutions for this equation are x = π/2 and x = 3π/2 when 0 ≤ x ≤ 2π.

The x-intercepts of f are the values of x for which f(x) = 0. From the previous question, we found that x = π/2 and x = 3π/2 are x-intercepts of f.