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Thread: Why isn't the other part of the graph taken?

  1. #1
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    Why isn't the other part of the graph taken?

    The question is-


    Make the function $f:R\rightarrow$$R$, $f(x)=x^2$ invertible by making suitable adjustments to the domain and range.


    So, here is what my teacher said-


    Since we have to make the function onto, we must make the range equal to co-domain.


    Range of y=$x^2$ is $[0\rightarrow \infty]$.


    Also, the function must be one-one. So, the domain of the function must be ahead of the minima of the function.


    So, finally the function becomes $f:[0\rightarrow \infty]\rightarrow [0 \rightarrow \infty]$.


    So, the graph looks like this

    [1]: https://i.stack.imgur.com/OkkHg.jpg



    But, why isn't this part of the graph also included in the answer

    [2]: https://i.stack.imgur.com/Y0S6R.jpg
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  2. #2
    MHF Contributor

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    Re: Why isn't the other part of the graph taken?

    It cannot be "included" in the answer because the two together give all of y= x^2 which is not invertible. Both -2 and 2 are mapped to 4 so there is not unique inverse for 4. However, that is an alternate answer. " y= x^2 for x\ge 0" has inverse y= \sqrt{x} while " y= x^2 for x\le 0" has inverse y= -\sqrt{x}. If fact, there are infinitely many correct answers to this problem. " y= x^2 for x> 1" is invertible and so is " y= x^2 for x< -2".
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