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Thread: How would I find the inverse of this function?

  1. #1
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    How would I find the inverse of this function?

    $\displaystyle f(x)=x^2-1$

    I know the first step is to change it to $\displaystyle y=x^2-1$ and then you would switch the x and y variables but when switching the x and y, do you just switch the variables in this case or when switching the variables do you include the $\displaystyle ^2$ of the x?

    so would it be $\displaystyle x^2=y-1$

    or
    $\displaystyle x=y^2-1$



    sorry, this is posted in the wrong section.
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  2. #2
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    Quote Originally Posted by Hypertension View Post
    $\displaystyle f(x)=x^2-1$

    I know the first step is to change it to $\displaystyle y=x^2-1$ and then you would switch the x and y variables but when switching the x and y, do you just switch the variables in this case or when switching the variables do you include the $\displaystyle ^2$ of the x?

    so would it be $\displaystyle x^2=y-1$

    or
    $\displaystyle x=y^2-1$



    sorry, this is posted in the wrong section.
    It's $\displaystyle x=y^2-1$.
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  3. #3
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    Quote Originally Posted by Hypertension View Post
    $\displaystyle f(x)=x^2-1$

    I know the first step is to change it to $\displaystyle y=x^2-1$ and then you would switch the x and y variables but when switching the x and y, do you just switch the variables in this case or when switching the variables do you include the $\displaystyle ^2$ of the x?

    so would it be $\displaystyle x^2=y-1$

    or
    $\displaystyle x=y^2-1$

    sorry, this is posted in the wrong section.
    Without the domain, where the function is defined, we cannot really tell if the function has an inverse.

    If $\displaystyle f:\mathbb{R} \to\mathbb{R} $, the the function has no inverse since the function is not one-one.

    To make it a bijective function, you have to first restrict the domain.

    I will give an example. Lets say $\displaystyle f:\mathbb{R}^+ \cup \{0\} \to [-1, \infty)$...

    Now its quite evident that f(x) is bijective, so the inverse function exists, call it g(x).

    By definition, to find the inverse function g(x), we need $\displaystyle g(f(x)) = f(g(x)) = x.$

    So let $\displaystyle g(x) = y$, then $\displaystyle f(g(x)) = x \Rightarrow f(y) = x \Rightarrow y^2 - 1 = x \Rightarrow y =\pm \sqrt{x + 1} $

    Now you will see why we need the bijection condition. As you can see the y that we have got is not a meaningful function. One of the basic properties of a function is that for 1 value of x there must be only one value of f(x).The second property is that it should be defined everywhere in the domain. But here we two values, $\displaystyle \sqrt{x + 1}$ and $\displaystyle -\sqrt{x + 1}$.

    This is where the domain helps. Assuming $\displaystyle \sqrt{x+1} = |\sqrt{x+1}| \geq 0$, and using the fact that the inverse function can only be in $\displaystyle \mathbb{R}^+ $, means $\displaystyle \sqrt{x+1}$ is the right choice.

    Secondly for $\displaystyle \sqrt{x+1}$ to be defined everywhere in the domain, the following should happen:

    $\displaystyle x+1 \geq 0 \Rightarrow x \geq -1 \Rightarrow x \in [-1,\infty)$

    So we see that $\displaystyle g: [-1, \infty) \to \mathbb{R}^+ \cup \{0\}, g(x) = \sqrt{x+1}$ is the inverse function.
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  4. #4
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    Quote Originally Posted by Hypertension View Post
    $\displaystyle f(x)=x^2-1$

    I know the first step is to change it to $\displaystyle y=x^2-1$ and then you would switch the x and y variables but when switching the x and y, do you just switch the variables in this case or when switching the variables do you include the $\displaystyle ^2$ of the x?

    so would it be $\displaystyle x^2=y-1$

    or
    $\displaystyle x=y^2-1$



    sorry, this is posted in the wrong section.
    You correctly said that you let the equation equal $\displaystyle y$. Following this, you make $\displaystyle x$ the subject and hence you will have the inverse. Remember to change your $\displaystyle y$ into $\displaystyle x$ in the end. It would be this:


    Let $\displaystyle y = x^2-1$

    $\displaystyle \therefore y + 1 = x^2$

    $\displaystyle \therefore x = \sqrt{y + 1}$

    $\displaystyle \implies f^{-1}(x) = \sqrt{x + 1}$
    Last edited by Simplicity; Jul 3rd 2008 at 09:32 AM.
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  5. #5
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    $\displaystyle f(x)=x^2-1$

    let
    $\displaystyle f(k)=x$
    $\displaystyle \therefore f^-1(x)=k$

    $\displaystyle f(k)=k^2-1 $
    $\displaystyle x=k^2-1 $ (because $\displaystyle f(k)=x$)
    $\displaystyle x+1=k^2$
    $\displaystyle k=\sqrt{x + 1}$
    $\displaystyle f^-1(x)=k$
    thus, $\displaystyle f^-1(x)=\sqrt{x + 1}
    $
    Last edited by z1llch; Jul 3rd 2008 at 07:42 AM. Reason: latex errrorrr
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  6. #6
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    Quote Originally Posted by z1llch View Post
    $\displaystyle f(x)=x^2-1$

    let
    $\displaystyle f(k)=x$
    $\displaystyle \therefore f'(x)=k$

    $\displaystyle f(k)=k^2-1 $
    $\displaystyle x=k^2-1 $ (because $\displaystyle f(k)=x$)
    $\displaystyle x+1=k^2$
    $\displaystyle k=\sqrt{x + 1}$
    $\displaystyle f'(x)=k$
    thus, $\displaystyle f'(x)=\sqrt{x + 1}
    $
    I see what you are doing here, but your notation could be easily confused with taking a derivative. I'd advise something like $\displaystyle f^{-1}$ rather than $\displaystyle f'$.

    -Dan
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    Thank you all.
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