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Math Help - Functions

  1. #1
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    Functions

    Hello

    The problem states:

    IF f(\frac{1}{x})=x^2+\frac{1}{x^3} AND g(x)= \frac{1}{2}(f(x)+f(-x)) THEN g(x)=?


    So, I got that f(x)=\frac{1}{x^2}+\frac{x^3}{1} and that f(-x)=\frac{1}{x^2}-\frac{x^3}{1}

    That gives me g(x)=\frac{1}{2}*\frac{2}{x^2}=\frac{1}{x^2}

    Is this correct? I had huge problems with this as I was always getting g(x)=\frac{1}{2}*0... then I figured my mistake: f(-x) is not equal to (-f(x))
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  2. #2
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    Awetuouncsygg
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    "Is this correct?" Yes, it is.
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  3. #3
    MHF Contributor

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    If f is any function, we can define the "even part" of f(x) to be f_e(x)= \frac{f(x)+ f(-x)}{2} and the "odd part" of f(x) to be [tex]f_o(x)= \frac{f(x)- f(-x)}{2}[/itex]. f_e is an "even function", f(-x)= f(x), f_o(x) is an "odd functin", f(-x)= -f(x), and their sum is f_e(x)+ f_o(x)= f(x).

    If f is an even function to start with, its "odd part" will be 0 and if f is an odd function, its "even part" will be 0. Rational functions that involve only even powers of x are "even functions" and rational functions that involve only odd powers of x are "odd functions.

    Here, you were asked to find the even part of the function f(x)= \frac{1}{x^2}+ x^3 which is, as you finally found, \frac{1}{x^2}. Its odd part is, of course, x^3.

    Here, you are basically asked
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