Let f(x) = 8sinx
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. The y-intercept of the graph of f is found by setting x = 0. Therefore, f(0) = 8sin(0) = 0. So, the y-intercept is at (0, 0).
b. The graph of f will be increasing when its derivative is positive. The derivative of f(x) = 8sin(x) is f'(x) = 8cos(x). The cosine function is positive in the interval -π/2 ≤ x ≤ π/2. Therefore, the graph of f is increasing when -π/2 < x < π/2.
c. The absolute minimum value of f occurs when sin(x) is at its minimum value of -1. So, the minimum value of f is f(x) = 8sin(x) = 8(-1) = -8.
d. To find the values of x that make f(x) = 0, we need to solve the equation 8sin(x) = 0. This occurs when sin(x) = 0. The solutions to sin(x) = 0 are x = 0, π, 2π. Therefore, for the interval 0 ≤ x ≤ 2π, f(x) = 0 at x = 0, π, and 2π.
e. The x-intercepts of f occur when f(x) = 0. So, the x-intercepts are the same as the solutions to sin(x) = 0, which are x = 0, π, and 2π. Therefore, the x-intercepts of f are at x = 0, π, and 2π.