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Math Help - Solving functions with graph, inverses and etc.?

  1. #1
    Member realintegerz's Avatar
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    Solving functions with graph, inverses and etc.?

    F(x) = { (3,5),(2,4),(1,7) }
    H(x) = { (3,2),(4,3),(1,6) }
    h(x) = sq. root (x-3)
    k(x) = (x^2) + 5

    Then I have a picture of the graph of f(x) and g(x) but not H & h

    I'm not sure how to do these:

    1/F(x), F^-1 (x), f^-1 (x)
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  2. #2
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    Quote Originally Posted by realintegerz View Post
    F(x) = { (3,5),(2,4),(1,7) }
    H(x) = { (3,2),(4,3),(1,6) }
    h(x) = sq. root (x-3)
    k(x) = (x^2) + 5

    Then I have a picture of the graph of f(x) and g(x) but not H & h

    I'm not sure how to do these:

    1/F(x), F^-1 (x), f^-1 (x)
    Could you provide us with the complete unabrigded and unedited text of your problem? What you have posted doesn't make sense to me ...
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  3. #3
    Senior Member Stroodle's Avatar
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    For the others you can swap the x and y over then solve for y. Also remember that the range of the original function becomes the domain of its inverse

    e.g.

    h(x)=\sqrt{x-3} , the range of this function is [0,\infty)

    y=\sqrt{x-3}

    x=\sqrt{y-3}

    y=x^2+3

    \therefore h^{-1}(x)=x^2+3 , the domain of this function is [0,\infty)
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  4. #4
    Member realintegerz's Avatar
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    Quote Originally Posted by earboth View Post
    Could you provide us with the complete unabrigded and unedited text of your problem? What you have posted doesn't make sense to me ...
    Well it is how it is intended to be..

    I got a graph with two functions on it, f and g, then I have those 4 equations for F(x), h(x), H(x), and k(x)

    I'm just having trouble understanding what those 3 mean...because F(x) is a set of values and f(x) I don't have the equation for
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  5. #5
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    Quote Originally Posted by realintegerz View Post
    Well it is how it is intended to be..

    I got a graph with two functions on it, f and g,

    ...
    Is it possible for you to post the drawing? (As you may know it is more than 1000 words )
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  6. #6
    Member realintegerz's Avatar
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    Oh I see what you mean... f and g are functions on a graph, I don't have the equations to them...

    All I have given is what I have above.
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  7. #7
    Member realintegerz's Avatar
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    That's what the problem looks like. Does it make a difference now?

    I tried to explain the best I could but now you got a picture if it helps.

    I still don't get how to solve these : 1/F(x), F^-1 (x), f^-1 (x)

    Since F is coordinates and f is on the graph.
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  8. #8
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    Talking

    To list the points in 1/F(x), take the x-values from the points in F(x), and invert the corresponding y-values. For instance, the point (3, 5) in F(x) becomes the point (3, 1/5) in 1/F(x).

    To list the points in the inverse of F(x), just apply the definition of the inverse and swap the x- and y-coordinates.

    The graph shows that f(x) is a linear equation, so use what you've learned about slope and straight-line equations to find the equation for f(x). Then follow the process you've learned for finding the inverse of a function.

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  9. #9
    Member realintegerz's Avatar
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    Ok so this is what I got:

    1/F(x) = {(3, 1/5), (2, 1/4), (1, 1/7)}

    F^-1(x) = {(5,3), (4, 2), (7, 1)}

    f^-1(x) = 1 - x
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