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, -2π ≤ x ≤ 2π​, does ​f(x)​ = 4?

e. What are the​ x-intercepts of​ f?

1 answer

(a) To find the y-intercept of the graph of f, we substitute x = 0 into the equation:
f(0) = 4cos(0) = 4(1) = 4
Therefore, the y-intercept of the graph of f is 4.

(b) In the given interval -π ≤ x ≤ π, the graph of f is decreasing when the cosine function is decreasing. The cosine function is decreasing in the third quadrant (where the x-values are greater than π/2 and less than π) and in the fourth quadrant (x-values greater than -π and less than -π/2). So, the graph of f is decreasing when x is between -π and -π/2, and also between π/2 and π.

(c) The absolute minimum value of f occurs when the cosine function is at its minimum value of -1. Therefore, the absolute minimum value of f is:
f(minimum) = 4cos(maximum) = 4(-1) = -4

(d) To find the values of x where f(x) = 4, we set the equation equal to 4 and solve for x:
4cos(x) = 4
cos(x) = 1
Since cosine function gives a maximum value of 1, we know that cos(x) = 1 when x is a multiple of 2π. Therefore, for -2π ≤ x ≤ 2π, the values of x where f(x) = 4 are x = 0, 2π, and -2π.

(e) The x-intercepts of f occur when the graph crosses the x-axis, meaning f(x) = 0. Therefore, we solve the equation:
4cos(x) = 0
cos(x) = 0
The values of x where cos(x) = 0 are multiples of π/2. So, the x-intercepts of f are x = π/2 and x = -π/2.