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Math Help - Functions and Inverses

  1. #1
    Junior Member Freaky-Person's Avatar
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    Functions and Inverses

    just 2. I got the rest down but these ones are driving me insane.

    Find inverse of functions

    1. h(x) = 2x/(x+10)

    2. f(x) = x^2 + 6x + 11

    Thanks.
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  2. #2
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    Quote Originally Posted by Freaky-Person View Post

    1. h(x) = 2x/(x+10)
    y=\frac{2x}{x+10}
    Substitute x for y and solve,
    x=\frac{2y}{y+10}
    x(y+10)=2y
    xy+10x=2y
    2y-xy=10x
    y(2-x)=10x
    y=\frac{10x}{2-x}
    2. f(x) = x^2 + 6x + 11
    It has no inverse.
    Horizontal line test, a parabola can be passes twice or more with a horizontal line.
    Last edited by ThePerfectHacker; January 14th 2007 at 09:02 PM.
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  3. #3
    Junior Member Freaky-Person's Avatar
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    It has no inverse.
    Horizontal line test, a parabola can be passes twice or more with a horizontal line.
    I know it stops being a function, but there should still be an inverse right? Cause you can still draw y=x and flip the original perabola along that line to get the shape, even if it's not a function.

    Like the inverse of y = X^2 is y = +-sqrtX, it's still an inverse even if it's not a function....
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    y=\frac{2x}{x+10}
    Substitute x for y and solve,
    x=\frac{2y}{y+10}
    x(y+10)=2y
    xy+10x=2y
    2y-xy=10x
    y(2-x)=10x
    y=\frac{10}{2-x}

    It has no inverse.
    Horizontal line test, a parabola can be passes twice or more with a horizontal line.
    Minor error on #1; did not carry over the x.

    Should be (10x)/(x-2)
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  5. #5
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    Quote Originally Posted by Freaky-Person View Post
    I know it stops being a function, but there should still be an inverse right? Cause you can still draw y=x and flip the original perabola along that line to get the shape, even if it's not a function.

    Like the inverse of y = X^2 is y = +-sqrtX, it's still an inverse even if it's not a function....
    f^(-1)(x) = {(y,x) | such that y = f(x)}

    f will not have an inverse if its not a one-to-one function.
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  6. #6
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by AfterShock View Post
    f^(-1)(x) = {(y,x) | such that y = f(x)}

    f will not have an inverse if its not a one-to-one function.
    whaaaa!! Then what am I supposed to do!?! It says
    For the function f(x) = x^2 + 6x + 11

    a)sketch the function and its inverse on the same grid

    b) Find the equation of the inverse


    ...
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  7. #7
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    Quote Originally Posted by Freaky-Person View Post
    Find inverse of functions

    2. f(x) = x^2 + 6x + 11
    Put:

    y=x^2+6x+11

    then:

    x^2+6x+11-y=0,

    Now use the quadratic formula to get:

    x=\frac{-6 \pm \sqrt{36-4(11-y)}}{2}

    or:

    g(y)=\frac{-6 \pm \sqrt{36-4(11-y)}}{2}.

    Which is not a function when you are looking for functions from
    R to R as it fails to be single valued, but it is what you are expected to
    produce, and there are interpretations under which it is a function
    (such as the extension of f and g to mappings from P(R) the set of
    subsets of R, to itself) .

    RonL
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  8. #8
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    Maybe he is trying to find the inverse,
    f^{-1}: f[\mathbb{R}]\to \mathbb{R}.
    Meaning from the image of the function to the set of real numbers. In that case an inverse exists.
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  9. #9
    Junior Member Freaky-Person's Avatar
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    Kay I got it. I'll post it up for future reference or something like that.

    f(x) = x^2 + 6x + 11

    y= x^2 + 6x + 11

    for inverse

    x = y^2 + 6y + 11

    x = (y^2 + 6y + 9) + 2

    x - 2 = (y + 3)^2

    +-sqrt(x - 2) = y + 3

    +-sqrt(x - 2) - 3 = y


    BTW, anyone know any sort of tutorial for that big letter CODE writing?
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  10. #10
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Maybe he is trying to find the inverse,
    f^{-1}: f[\mathbb{R}]\to \mathbb{R}.
    Meaning from the image of the function to the set of real numbers. In that case an inverse exists.
    I have no idea what you just said there, but it's probably it.
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  11. #11
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    Quote Originally Posted by Freaky-Person View Post

    BTW, anyone know any sort of tutorial for that big letter CODE writing?
    I am impressed, a teenager who actually wants to learn something! I wish more of your types existed, otherwise their philosphy is simple, "If it not on the exam we do not care, we do not need it".
    http://www.mathhelpforum.com/math-he...-tutorial.html
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