MATH SOLVE

4 months ago

Q:
# Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = 8 cos2(x) − 16 sin(x), 0 ≤ x ≤ 2π (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Correct: Your answer is correct. Find the interval on which f is decreasing. (Enter your answer using interval notation.) Correct: Your answer is correct. (b) Find the local minimum and maximum values of f. local minimum value Correct: Your answer is correct. local maximum value Correct: Your answer is correct.

Accepted Solution

A:

Answer:increasing: (π/2, 3π/2)decreasing: [0, π/2) ∪ (3π/2, 2π]minimum: -16 at x=π/2maximum: 16 at x=3π/2Step-by-step explanation:If all you want are answers to the questions, a graphing calculator can provide them quickly and easily. (see attached)___If you need an algebraic solution, you need to find the zeros of the derivative. f'(x) = -16cos(x)sin(x) -16cos(x) = -16cos(x)(sin(x) +1)The product is zero where the factors are zero, at x=π/2 and x=3π/2.These are the turning points, where the function changes from decreasing to increasing and vice versa.(sin(x)+1) is non-negative everywhere, so the sign of the derivative is the opposite of the sign of the cosine function. This tells us the function f(x) is increasing on the interval (π/2, 3π/2), and decreasing elsewhere (except where the derivative is zero).The function local extrema will be where the derivative is zero, so at f(π/2) (minimum) and f(3π/2) (maximum). We already know that cos(x) is zero there, so the extremes match those of -16sin(x).