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.)

See attached photos, for some reason, it won't let me submit an answer, but I showed all the work and gave explanations in the photos

Step-by-step explanation:

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

because they could drop even lower..