Find interval of increase and decrease of f(x) = 8 sin(x) + cot(x), −π ≤ x ≤ π

Given f(x)=8sin(x)+cot(x) for -pi<x<pi :

Note that:

f'(x)=8cos(x)-csc^2(x)

f''(x)=-8sin(x)+2csc^2(x)cot(x)

(1) To find the intervals where f(x) is increasing or decreasing we use the first derivative test; if the first derivative is positive on an interval the functio is increasing, negative implies the functio is decreasing.

Using technology we find the approximate zeros of f'(x) on -pi<x<pi :

x~~-1.443401

x~~-.3752857

x~~.3752857

x~~1.443401

Plugging in test values on the intervals yields:

f'(x)<0 on (-pi,-1.443401)

f'(x)>0 on (-1.443401,-.3752857)

f'(x)<0 on

Plz correct me if wrong

the answer is: the length is 5/6 yards.

give me a or a 5 star rating if this is correct!

step-by-step explanation:

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!