Produce graphs of f that reveal all the important aspects of the curve. then use calculus to find the following. (enter your answers using interval notation. round your answers to two decimal places.)
f(x) = 3 sin x + cot x, -

a) find interval of decrease
b)find interval of increase
c)find inflection points of the function and their y values
d) find interval where the function is concave up; where f is concave down

Step-by-step explanation:

given that the function is given from the question as;

f(x) = 3 sin x + cot x, where -π ≤ x ≤ π

solving;

given f(x) = 3 sin x + cot x

f(x) = 3 cos x - cos²x = 0

which is 3 cos x - 1/sin²x = 0

3 cos x = 1/sin²x

∴ 3 cosx.sin²x = 1

simplifying gives, 3cosx(1-cos²x) = 1

expanding gives 3cos³x - 3cosx + 1 = 0

given x = -1.165, -0.734, 0.734 , 1.165

here x is critical at 0

for region 1; (-π, -1.165)

for region 2; (-1.165, -0.734) ⇒ increases

for region 3; (-0.734, 0) ⇒ decreases

for region 4; (0, 0.734) ⇒ decreases

for region 5; (0.734, 1.165) ⇒ increases

for region 6; (1.165,π) ⇒ decreases

(a). interval of decrease becomes : (π,-1.165)U(-0.734,0)U(0,0.734)U(1.165,π)

(b). interval of increase becomes : (-1.165,-0.734)U(0.734,1.165)

(c). to solve for the inflection points, we input the equation

f(x) = sin x (2 cotx cos³x -3)

inputting we have that

x = -0.922 and 0.922

x = 0

for region 1; (-π, -0.922) ⇒ concave up

for region 2; (-0.922, 0) ⇒ concave down

for region 3; (0,0.922) ⇒ concave up

for region 4; (0.922,π) ⇒ concave down

(d). For concave up; (-π,-0.922)U(0,0.922)

     For concave down; (-0.922,0)U(0.992, π)

cheers i hope this helps

fdvs

step-by-step explanation: vsfdvsfvd