Question

For y=(1/2)x-sinx where 0 15 years ago

Answer 1

To find the domain, all you care about is

1), is the denominator = 0 ?

2), is the number inside a square root negative ?

Answer 2

in your case, x/2 is a polynomial and sin(x) is a trig function
with no vertical asymptotes, so the denominator is never 0
and we don't have to worry about imaginary numbers.

Answer 3

trig functions oscilate, so there is no need to consider
about horizontal asymptotes.

Answer 4

for the same reason, slant asymptotes will not be there

Answer 5

For domain it is provided, 0<x<3pi
For vertical asymptotes look for x-values within domain that make y undefined, in this case y is defined on all of x
Horizontal asymptotes can be obtained by looking at the limit of function as x->infinity, but in this case with a trig function there is no limit because its cyclical.
x-intercept, set y=0
sinx = 1/2x

Answer 6

vertical asymptotes are the points where your function has a
denominator equal to zero, but not the numerator.
again, since there were not problem with the domain
we don't have to worry about that.

Answer 7

for the x-int, solve for x when y=0
for the y-int, solve for y when x=0.

Answer 8

to see whether f is increasing or decreasing,
you will find the derivative of your function f.

if f'>0, then the function is increasing.
the opposite would work similarly.

Answer 9

to see whether f is concave up or down,
you will find the second derivative of your function f.

if f">0, then it is concave up.
the opposite is similar.

Answer 10

let me know if you need more help.

Answer 11

Wow, thanks a lot!