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anonymous

  • 5 years ago

For y=(1/2)x-sinx where 0<x<3pi How do I get the Domain, Horizontal Asymptotes, Vertical Asymptotes, Slant asymptotes, x and y intercepts, increasing and decreasing intervals, as well the concativity and inflection points. How do you get all of that in cases like this one where the equation is trigonometric?

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  1. Yuki
    • 5 years ago
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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 ?

  2. Yuki
    • 5 years ago
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    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.

  3. Yuki
    • 5 years ago
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    trig functions oscilate, so there is no need to consider about horizontal asymptotes.

  4. Yuki
    • 5 years ago
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    for the same reason, slant asymptotes will not be there

  5. dumbcow
    • 5 years ago
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    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

  6. Yuki
    • 5 years ago
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    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.

  7. Yuki
    • 5 years ago
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    for the x-int, solve for x when y=0 for the y-int, solve for y when x=0.

  8. Yuki
    • 5 years ago
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    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.

  9. Yuki
    • 5 years ago
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    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.

  10. Yuki
    • 5 years ago
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    let me know if you need more help.

  11. anonymous
    • 5 years ago
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    Wow, thanks a lot!

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