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Math Help - Functions

  1. #1
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    Functions

    1. The function f(x) is not defined for x = 0. It has the property that for all nonzero real numbers x, f(x) + 2f(1/x) = 3x. Find all values of a such that f(a) = f(-a)

    2. The function f is defined by f(x) = (ax+b)/(cx+d), where a, b, c, and d are nonzero real numbers, and has the properties: f(19) = 19, f(97) = 97, and f(f(x) = x for all values of x except -d/c. Find the unique number that is not in the range of f.

    First time seeing this. Somehow tell me how to approach it, I don't need the answer.
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  2. #2
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    2.
    Since we are given \frac{-d}{c} isn't in the domain, we know that any number of the form \frac{-ad}{c}+b isn't in the range of f.
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  3. #3
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    For the first one, try substituting a in for x, and then 1/a in for x. Solve the resulting system of equations for f(a) or f(1/a) and see what pops out.
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  4. #4
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    Rational function.

    Quote Originally Posted by awkwardnerd View Post
    1. The function f(x) is not defined for x = 0. It has the property that for all nonzero real numbers x, f(x) + 2f(1/x) = 3x. Find all values of a such that f(a) = f(-a)

    2. The function f is defined by f(x) = (ax+b)/(cx+d), where a, b, c, and d are nonzero real numbers, and has the properties: f(19) = 19, f(97) = 97, and f(f(x) = x for all values of x except -d/c. Find the unique number that is not in the range of f.

    First time seeing this. Somehow tell me how to approach it, I don't need the answer.
    #2.
    \displaystyle f(f(x))=x\quad\implies\quad f^{-1}(x)=f(x).

    \displaystyle f(x) has a vertical asymptote at \displaystyle x=-\,{{d}\over{c}} and a horizontal asymptote at \displaystyle f(x)={{a}\over{c}}. Therefore, \displaystyle f^{-1}(x) has a horizontal asymptote at \displaystyle x=-\,{{d}\over{c}} and a vertical asymptote at \displaystyle f^{-1}(x)={{a}\over{c}}.

    Thus \displaystyle f^{-1}(x)=f(x)\quad\implies\quad -\,{{d}\over{c}}={{a}\over{c}} so that \displaystyle a=-d.

    Also, the zero of \displaystyle f(x) is equal to the y-intercept of \displaystyle f^{-1}(x), and vice-versa, but I see that this just gives \displaystyle d=-a again.

    So, we have that \displaystyle f(x)={{ax+b}\over{cx-a}}.

    We also know that \displaystyle f(x)=x for \displaystyle x=19,\ 97\,. Solve \displaystyle f(x)=x and that should give us \displaystyle x=58\pm39=97,\ 19\,.

    \displaystyle {{ax+b}\over{cx-a}}=x

    \displaystyle ax+b=x(cx-a)

    \displaystyle cx^2-2ax-b=0

    \displaystyle x={{2a\pm\sqrt{4a^2+4bc}}\over{2c}}={{a\pm\sqrt{a^  2+bc}}\over{c}} ={{a}\over{c}}\pm{{\sqrt{a^2+bc}}\over{c}}=58\pm39

    Thus, \displaystyle a=58c and \displaystyle {{\sqrt{a^2+bc}}\over{c}}=39\ \ \implies\ \ b=(39^2-58^2)c \ \implies\ \ b=-(19)(97)c

    Therefore, \displaystyle f(x)={{58cx-(19)(97)c}\over{cx-58c}}. Cancel \displaystyle c.

    \displaystyle f(x)={{58x-1843}\over{x-58}}.

    Last edited by SammyS; December 17th 2010 at 10:52 PM. Reason: Additional material.
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  5. #5
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    Quote Originally Posted by DrSteve View Post
    For the first one, try substituting a in for x, and then 1/a in for x. Solve the resulting system of equations for f(a) or f(1/a) and see what pops out.
    Plugging in a for x: f(a)+2f(\frac{1}{a})=3a
    Pulgging in \frac{1}{a} for x: f(\frac{1}{a})+2f(a)=\frac{3}{a}

    Solving both of these for f(a):
    f(a)=3a-2f(\frac{1}{a})=\frac{3}{2a}-\frac{f(\frac{1}{a})}{2}
    Also, f(a)=f(-a) \quad\implies\quad -3a-2f(\frac{-1}{a})=\frac{-3}{2a}-\frac{f(\frac{-1}{a})}{2}
    And -3a-2f(\frac{1}{a})=\frac{-3}{2a}-\frac{f(\frac{1}{a})}{2}

    When I look at these all I get is a=0, which is excluded from the domain. I get the same thing when I solve for f(\frac{1}{a}). What am I missing?
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  6. #6
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    Quote Originally Posted by awkwardnerd View Post
    1. The function f(x) is not defined for x = 0. It has the property that for all nonzero real numbers x, f(x) + 2f(1/x) = 3x. Find all values of a such that f(a) = f(-a)

    First time seeing this. Somehow tell me how to approach it, I don't need the answer.

    #1.
    pflo's post gave me the idea for the solution.

    \displaystyle f(x)+2f\left({{1}\over{x}}\right)=3x\ \ \to\ \ -f(x)-2f\left({{1}\over{x}}\right)=-3x

    \displaystyle f\left({{1}\over{x}}\right)+2f(x)={{3}\over{x}}\ \ \ \to\ \ \ 4f(x)+2f\left({{1}\over{x}}\right)=\ {{6}\over{x}}

    Add the equations on the right to eliminate f\left({{1}\over{x}}\right), then solve for \displaystyle f(x).

    As it turns out, f is an odd function, so \displaystyle f(a)=f(-a) implies that \displaystyle f(a)=0.

    There are two values of a which satisfy this condition.
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  7. #7
    Member pflo's Avatar
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    Quote Originally Posted by SammyS View Post
    As it turns out, f is an odd function, so \displaystyle f(a)=f(-a) implies that \displaystyle f(a)=0.
    I agree that f is an odd function because f(-a)=f(a), but why would this imply that f(a)=0?
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  8. #8
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    Quote Originally Posted by pflo View Post
    I agree that f is an odd function because f(-a)=f(a), but why would this imply that f(a)=0?
    No, the definition of odd function is that f(-a)= -f(a). If you know that f is an odd function and that f(-a)= f(a) then f(a)= -f(a) so that f(a)= 0.
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  9. #9
    Member pflo's Avatar
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    Duh. Thanks. I wasn't thinking straight. If f(-x)=f(x), then f is an even function! But this isn't always true for this function: we're just looking for values of x where this would be true.

    So this brings me to another question: how do we know that the function at issue here is an odd function?
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  10. #10
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    Note that f being odd always implies that f(0) = 0 (you don't need the other condition). Indeed, if f(-x)=-f(x) for all x, then substituting a 0 in for x yields f(-0)=-f(0). But -0 = 0 whence f(0) = -f(0). Adding f(0) to both sided yields 2f(0) = 0. Finally, divide both sides by 2 to get f(0) = 0.

    Edit: I should clarify that f being odd always implies that f(0)=0 whenever 0 is in the domain of f.
    Last edited by DrSteve; December 21st 2010 at 07:13 AM.
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  11. #11
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    Here is how we know that f is odd:

    Adding the two equations in Sammy's post shows that 3f(x)=-3f(x)+\frac{6}{x}, so that f(x)=-f(x)+\frac{2}{x}. Thus, 2f(x) = \frac{2}{x} , and f(x) = \frac{1}{x} which is odd.
    Last edited by DrSteve; December 21st 2010 at 07:15 AM.
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