Find the equilibrium point, the maximum and the minimum values for the the following trigonometric function. Sketch the graph in your notebook.𝑓(𝑥)=4cos(4𝑥)+4

Sure, let's break this down step by step.

1. Equilibrium Point: The equilibrium point of a function is the x-value where the function equals its average value. For a cosine function, the average value is the vertical shift, which in this case is 4. So, the equilibrium point is any x such that f(x) = 4. Since cos(4x) oscillates between -1 and 1, f(x) = 4cos(4x) + 4 oscillates between 3 and 5. Therefore, the equilibrium point is any x such that 4cos(4x) + 4 = 4, which simplifies to cos(4x) = 0. The solutions to this are x = (2n+1)π/8 for any integer n.

2. Maximum and Minimum Values: The maximum and minimum values of a cosine function are given by the amplitude plus/minus the vertical shift. The amplitude of f(x) is 4 (the coefficient of the cosine function), and the vertical shift is also 4. Therefore, the maximum value is 4 + 4 = 8, and the minimum value is 4 - 4 = 0.

3. Sketching the Graph: To sketch the graph of f(x) = 4cos(4x) + 4, start by drawing the x-axis and y-axis. Mark the equilibrium point on the y-axis at y = 4. Then, draw a wave that oscillates between the maximum and minimum values (8 and 0), with a period of 2π/4 = π/2. The wave should cross the x-axis at the equilibrium points, which are (2n+1)π/8 for any integer n.

Please note that I can't physically draw the graph in a notebook for you as I'm an AI, but I hope these instructions help you to do it yourself!