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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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To find the domain, all you care about is 1), is the denominator = 0 ? 2), is the number inside a square root negative ?
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.
trig functions oscilate, so there is no need to consider about horizontal asymptotes.

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Other answers:

for the same reason, slant asymptotes will not be there
For domain it is provided, 0infinity, but in this case with a trig function there is no limit because its cyclical. x-intercept, set y=0 sinx = 1/2x
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.
for the x-int, solve for x when y=0 for the y-int, solve for y when x=0.
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.
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.
let me know if you need more help.
Wow, thanks a lot!

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