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Thread: 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 $\displaystyle f(x)$ as $\displaystyle x \rightarrow f(x)$, then $\displaystyle f(x)^{-1}$ is $\displaystyle f(x) \rightarrow x$.
    The first one is pretty straightforward : you are asked to find $\displaystyle f(1)^{-1}$ when $\displaystyle f(x) = x^2$.

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

    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 $\displaystyle f(x)$ as $\displaystyle x \rightarrow f(x)$, then $\displaystyle f(x)^{-1}$ is $\displaystyle f(x) \rightarrow x$.
    The first one is pretty straightforward : you are asked to find $\displaystyle f(1)^{-1}$ when $\displaystyle f(x) = x^2$.

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

    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 $\displaystyle (0,1)$ by $\displaystyle f^{-1}$
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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 $\displaystyle (0,1)$ by $\displaystyle f^{-1}$
    and that depends on $\displaystyle f^{-1}$,
    if $\displaystyle f^{-1}$ is increasing the interval image must be $\displaystyle (f^{-1}(0),f^{-1}(1))$.
    if $\displaystyle f^{-1} $is decreasing the interval must be $\displaystyle (f^{-1}(1),f^{-1}(0))$.
    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 $\displaystyle f^{-1}$,
    if $\displaystyle f^{-1}$ is increasing the interval image must be $\displaystyle (f^{-1}(0),f^{-1}(1))$.
    if $\displaystyle f^{-1} $is decreasing the interval must be $\displaystyle (f^{-1}(1),f^{-1}(0))$.
    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 $\displaystyle g(x)= \left |x \right |$ ?
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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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