Please answer the following Let f (r) sin on the interval [0,2m). Fill in the bl
ID: 2848269 • Letter: P
Question
Please answer the following
Explanation / Answer
f(x) = x + sinx
f'(x) = 1+cosx
1 + cosx is always greater than or equal to 0
1.so f(x) is always increasing.
2/so f(x) increasing for [0,2pi]
f(x) decreasing for none.
f'(x) =0
=> cosx = -1
=> x=pi.
at x=pi
f(x) = pi + sin(pi) = pi.
3.so local minima occurs at x=pi value of f(x) =pi.
4.there is no local maxima.
5.absolute maximum of f(x) occurs at x =2pi.
it's value = 2pi + sin(2pi) = 2pi
6.absolue minima occurs at x =0
f(x) = 0 + sin(0) = 0.
7.concave up for [0,2pi]
8. concvae down for none
9.points of infelction occur at x = pi.
10.no vertical assymptote.
11. y intercept at y=0