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Math Help - inverse function questions

  1. #1
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    inverse function questions

    Hi, I want to learn inverse functions. And I find a problem can you help me to solve this problems ?
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  2. #2
    Super Member Bacterius's Avatar
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    Basically, if you have f(x) as x \rightarrow f(x), then f(x)^{-1} is f(x) \rightarrow x.
    The first one is pretty straightforward : you are asked to find f(1)^{-1} when f(x) = x^2.

    Say f(x) = y, you have y = x^2. Express x in terms of y, thus x = \sqrt{y} or x = - \sqrt{y}. Therefore, f(x)^{-1} = \sqrt{x} or - \sqrt{x}. Substitute : f(1)^{-1} = 1 or -1

    I don't get b), though, what do you mean by the interval ?
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  3. #3
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    Quote Originally Posted by Bacterius View Post
    Basically, if you have f(x) as x \rightarrow f(x), then f(x)^{-1} is f(x) \rightarrow x.
    The first one is pretty straightforward : you are asked to find f(1)^{-1} when f(x) = x^2.

    Say f(x) = y, you have y = x^2. Express x in terms of y, thus x = \sqrt{y} or x = - \sqrt{y}. Therefore, f(x)^{-1} = \sqrt{x} or - \sqrt{x}. Substitute : f(1)^{-1} = 1 or -1

    I don't get b), though, what do you mean by the interval ?
    I don't know I research everywhere but I didn't find anything about interval in inverse function
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  4. #4
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    hi
    for (b) u should find the interval image of (0,1) by f^{-1}
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  5. #5
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    Quote Originally Posted by Raoh View Post
    hi
    for (b) u should find the interval image of (0,1) by f^{-1}
    and that depends on f^{-1},
    if f^{-1} is increasing the interval image must be (f^{-1}(0),f^{-1}(1)).
    if f^{-1} is decreasing the interval must be  (f^{-1}(1),f^{-1}(0)).
    hope that's right..
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  6. #6
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    Quote Originally Posted by Raoh View Post
    and that depends on f^{-1},
    if f^{-1} is increasing the interval image must be (f^{-1}(0),f^{-1}(1)).
    if f^{-1} is decreasing the interval must be  (f^{-1}(1),f^{-1}(0)).
    hope that's right..
    thank you Raoh , can you help for ii) ?
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  7. #7
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    for ii) is it g(x)= \left |x  \right | ?
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  8. #8
    Super Member Bacterius's Avatar
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    I don't think so, his function is the floor function, not the absolute value function ...
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