Analyze the trigonometric function f(x) = (4sinx + 4cosx)^2 over the interval x(-pie,pie), stating where f is increasing, decreasing, concave up, concave down, and stating the x-coordinates of all inflection points. A) Find the interval(s) on which f(x) is increasing choose: f(x) is never increasing 1 2 3 B) Find the Interval(s) on which f(x) is decreasing choose: f(x) is never idecreasing 1 2 3 c) Find the Interval(s) on which f(x) is concave up choose: f(x) is never concave up 1 2 3 d) FInd the interval(s) on which f(x) is concave down choose: f(x) is never concave down 1 2 3 e) Find the x-coordinate(s) of any/all inflection point(s). choose: f(x) has no inflection point 1 2 3

To answer this question, we first need to understand the function and its derivatives. The function given is f(x) = (4sinx + 4cosx)^2.

A) To find where the function is increasing, we need to find its first derivative and set it greater than 0. The first derivative of f(x) is f'(x) = 8(sin(x) + cos(x))(cos(x) - sin(x)). Solving f'(x) > 0, we get the intervals (-π/4, π/4) and (3π/4, 5π/4). So, the answer is 3.

B) Similarly, to find where the function is decreasing, we set the first derivative less than 0. Solving f'(x) < 0, we get the intervals (π/4, 3π/4) and (5π/4, 7π/4). So, the answer is 3.

C) To find where the function is concave up, we need to find its second derivative and set it greater than 0. The second derivative of f(x) is f''(x) = -8(sin(2x) + 2). Solving f''(x) > 0, we get the intervals (-3π/4, -π/4) and (π/4, 3π/4). So, the answer is 3.

D) Similarly, to find where the function is concave down, we set the second derivative less than 0. Solving f''(x) < 0, we get the intervals (-π, -3π/4), (-π/4, π/4), (3π/4, π). So, the answer is 3.

E) The inflection points are the points where the second derivative changes sign. From the intervals found in parts C and D, we can see that the inflection points are at x = -3π/4, -π/4, π/4, 3π/4. So, the answer is 3.