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Math Help - summer work

  1. #1
    Junior Member
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    Aug 2009
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    Talking summer work

    I'm working on a summer assignment for AP Calculus and am about three fourths done, just a couple things I seemed to have forgotten. review excercises for the course itself so it has some some content that isn't really calculus but it's work I got for a calculus class so i put it here if you guys don't mind .

    1. y=1-cosx
    2. y= (x^4 + 1) / (x^3 - 2x)
    3. y=x+cosx

    ... I know that if you plug in -x into the equations and get the original equation, it's even, and if you get the opposite, it's odd and if it's neither the equation is neither... but I've completely forgotten how to apply this when it comes to trigonoemetric functions.
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  2. #2
    Super Member
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    Quote Originally Posted by fezz349 View Post
    I'm working on a summer assignment for AP Calculus and am about three fourths done, just a couple things I seemed to have forgotten. review excercises for the course itself so it has some some content that isn't really calculus but it's work I got for a calculus class so i put it here if you guys don't mind .

    1. y=1-cosx
    2. y= (x^4 + 1) / (x^3 - 2x)
    3. y=x+cosx

    ... I know that if you plug in -x into the equations and get the original equation, it's even, and if you get the opposite, it's odd and if it's neither the equation is neither... but I've completely forgotten how to apply this when it comes to trigonoemetric functions.
    What do you want to do with these? Solve for x? Prove they are even/odd?

    If you want to prove they are even/odd:

    f(x) is even iff f(-x) = f(x)
    f(x) is odd iff f(-x) = -f(x)

    (I) Let f(x) = 1-cos(x), then: f(-x) = 1-cos(-x) = 1-cos(x) since cos(x) is an even function, so cos(-x)=cos(x) thus f(x) is even as well.

    (II) Let g(x) = \frac{x^4+1}{x^3-2x} then: g(-x) = \frac{x^4+1}{(-x)^3+2x} = \frac{x^4+1}{-(x^3-2x)} = -g(x), so g is odd.

    (III) Let h(x) = x + cos(x) then: h(-x) = -x + cos(-x) = -x+cos(x) so h(x) is neither even nor odd.
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