Consider the equation below. (if an answer does not exist, enter dne.) f(x) = 3 cos2(x) − 6 sin(x), 0 ≤ x ≤ 2π (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the interval on which f is decreasing. (enter your answer using interval notation.) (b) find the local minimum and maximum values of f. local minimum value local maximum value (c) find the inflection points. (x, y) = (smaller x-value) (x, y) = (larger x-value) find the interval on which f is concave up. (enter your answer using interval notation.) find the interval on which f is concave down. (enter your answer using interval notation.) need ? read it watch it

f(x) = 3 cos2(x) − 6 sin(x), 0 ≤ x ≤ 2π

We plot the function using a graphing calculator (See attached image below)

a) 1. Find the interval on which f is increasing.

f(x) increases for

x ∈ [π/2 , 7π/6] ∪ [3π/2, 11π/6]

2. Find the interval on which f is decreasing

f(x) decreases for

x ∈ [0 , π/2) ∪ (7π/6, 3π/2) ∪ (11π/6, 2π]

(b) Find the local minimum and maximum values of f

Local minimum.      f = -9

Local maximum.      f = 4.5

(c) 1. Find the inflection points

Please see second image attached

Points,

(x,y) = (2.50673, -2.66882)

(x,y) = (4.14456, 3.79382)

(x,y) = (5.28022, 3.79382)

2. Find the interval on which f is concave up.

x ∈ [0 , 2.50673] ∪ [4.14456, 5.28022]

3. Find the interval on which f is concave down.

x ∈ (2.50673, 4.14456) ∪ (5.28022, 2π]

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2