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Math Help - How would I find the inverse of this function?

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

    f(x)=x^2-1

    I know the first step is to change it to  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 ^2 of the x?

    so would it be x^2=y-1

    or
    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
    f(x)=x^2-1

    I know the first step is to change it to  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 ^2 of the x?

    so would it be x^2=y-1

    or
    x=y^2-1



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

    I know the first step is to change it to  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 ^2 of the x?

    so would it be x^2=y-1

    or
    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 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 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 g(f(x)) = f(g(x)) = x.

    So let g(x) = y, then 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,  \sqrt{x + 1} and -\sqrt{x + 1}.

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

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

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

    So we see that 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
    f(x)=x^2-1

    I know the first step is to change it to  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 ^2 of the x?

    so would it be x^2=y-1

    or
    x=y^2-1



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


    Let y = x^2-1

    \therefore y + 1 = x^2

    \therefore x = \sqrt{y + 1}

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

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

    f(k)=k^2-1
    x=k^2-1 (because f(k)=x)
    x+1=k^2
    k=\sqrt{x + 1}
    f^-1(x)=k
    thus, f^-1(x)=\sqrt{x + 1}<br />
    Last edited by z1llch; July 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
    f(x)=x^2-1

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

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

    -Dan
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    Thank you all.
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