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Math Help - inverse & rational functions

  1. #1
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    inverse & rational functions

    Please help!
    I have the solutions to them, I am just trying to understand how they were reached.
    Find the inverse of the given functions and find the domain for f(x) and f^-1(x)
    so I understand that you just flip the x and y
    but what does it mean domain?
    heres a couple
    f(x)=x-2 over x+1
    I know dom f(x)= [-2, infinity) but how is dom f^-1(x) = x+2 over 1-x??
    maybe that explanation will help me with the other ones

    rational functions:
    heres a couple, I understand how to do most of them but I get stuck at little aspects of it.
    find domain, intercepts, and asymptotes when appropriate:
    y=(x-2)^2 + 1
    why is there no x intercept? I got it as being x= square root of 5

    y= 1 over x-1 +2
    (the 2 is separate from the fraction)
    How is that =0 solved? for some reason I got x=0, I guess its because I don't know how to solve that equation since the x is in denominator. but the solution says x=1/2

    Final is tomorrow:/
    Thanks so much for all your help!
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  2. #2
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    If f^{-1}(x) is the inverse function of f(x), then the domain of f^{-1}(x) is the range of f(x) and the range of f^{-1}(x) is the domain of f(x). Like the variables x and y are 'flipped' to find the inverse function, the domain and range are also 'flipped'.



    The graph of y = (x - 2)^2 + 1 crosses the x-axis if and only if y = 0. However, y never equals 0 because the discriminant of y is negative. To review, the discriminant of ax^2 + bx + c is given by b^2 - 4ac.

    You may recall the discriminant reveals the number of real solutions to a quadratic equation. If the disciminant is positive, there are 2 real solutions and the graph crosses the x-axis twice. If the discriminant is zero, there is one real solution and the graph crosses the x-axis once. If the discriminant is negative, there are no real solutions and the graph does not cross the x-axis.

    Before you can find the discriminant of y, you have to expand y. For example, y = (x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5.



    To find the x-intercept(s) of y = \frac{1}{x - 1} + 2, substitute y = 0 and solve for x.

    0 = \frac{1}{x - 1} + 2

    Multiply both sides of the equation by x - 1.

    (x - 1) \times 0 = (x - 1) \times \frac{1}{x - 1} + 2

    Anything times 0 is still 0.

    0 = (x - 1) \times (\frac{1}{x - 1} + 2)

    Distribute (x - 1) into (\frac{1}{x - 1} + 2).

    0 = (x - 1) \times \frac{1}{x - 1} + (x - 1) \times 2

    (x - 1) and \frac{1}{x - 1} cancel, producing 1.

    0 = 1 + 2(x - 1)

    Subtract 1 from both sides of the equation.

    -1 = 2(x - 1)

    Divide by 2 on both sides of the equation.

    \frac{-1}{2} = x - 1

    Add 1 to both sides of the equation.

    \frac{-1}{2} + 1 = x

    Find the least common denominator of \frac{-1}{2} and 1.

    \frac{-1}{2} + \frac{2}{2} = x

    Write the left side of the equation as only one fraction.

    \frac{-1 + 2}{2} = x

    Simplify the numerator.

    \frac{1}{2} = x
    Last edited by NOX Andrew; December 15th 2009 at 07:15 PM.
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  3. #3
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    im sorry, I'm still confused. Is there anyway you (or anyone) could walk me through how to solve the whole problem?
    Here is another problem if this would be easier=
    f(x)= the square root of (x+2)
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  4. #4
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    If by solve f(x) = \sqrt{x + 2} you mean find the x-intercept(s), then set f(x) to 0 and solve for x.

    f(x) = 0

    Substitute f(x) = \sqrt{x + 2}.

    \sqrt{x + 2} = 0

    Square both sides of the equation.

    (\sqrt{x + 2})^2 = 0^2

    On the left side of the equation, the square root and the square cancel. On the right side of the equation, 0 squared is still 0.

    x + 2 = 0

    Subtract two from both sides of the equation.

    x = -2

    By the way, what confused you in my previous reply? I may be able to clarify it.
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  5. #5
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    O im sorry! when I saw the first post I didn't see the second part, I meant I am still confused about the inverse functions so that is what that problem is for. but I just read over the 2nd part of that post and that clears rational functions up! thank you soooo much
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  6. #6
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    I'll try to clarify what I wrote on inverse functions.

    You already know how to find the inverse of a function: flip the variables.

    To find the domain and range of an inverse function, flip the domain and range of the original function.

    For example, if the domain of a function is (-3,3) and the range of the function is (0,10), then the domain of the inverse is (0,10) and the range of the inverse is (-3,3). Do you see how the domain and range flipped? It's as easy as finding the domain and range of the original function and flipping them.
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  7. #7
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    I understand that concept, but how would I find the range for lets say
    (1/x)-1
    b/c the domain is all real numbers correct?
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