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$g(x)=4-3\sin x$ is defined for the domain $x \in [\pi/2,A].$ The question is to find the largest value of $A$ for which $g$ has an inverse. I was thinking to find the inverse of $g(x)$, and find the range of $g(x)$ since the range of the inverse=domain of the original function, but I didn't get the right answer.

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    $\begingroup$ It's better to do it generally: When does a function not have an inverse function? $\endgroup$
    – Arthur
    May 2 '17 at 13:46
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    $\begingroup$ @Arthur when it's one on one? $\endgroup$
    – JohnFire
    May 2 '17 at 13:52
  • $\begingroup$ @JohnFire It needs to be onto too. $\endgroup$ May 2 '17 at 13:54
  • $\begingroup$ @JohnFire Here is mathjax tutorial to format equations. $\endgroup$
    – Arbuja
    May 2 '17 at 13:58
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    $\begingroup$ Should be $(3/2) pi$ then @Arthur $\endgroup$
    – JohnFire
    May 2 '17 at 14:02
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The inverse is intended to be a function. This means it cannot take on multiple values as the "vertical line test." What this means for $g(x)$ is that it cannot take on multiple values as a "horizontal line test" otherwise its inverse $g^{-1}(x)$ would fail the vertical line test. Working through this this will give you the interval.

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  • $\begingroup$ Oh, so one x value can only have one y value, and only then we can have an inverse? $\endgroup$
    – JohnFire
    May 2 '17 at 13:51
  • $\begingroup$ One x value can only have one y value is what makes it a function. One y value can only have one x value is what makes it have an inverse. $\endgroup$ May 2 '17 at 14:02
  • $\begingroup$ of fives Does that mean $x^2$ is not a function then? $\endgroup$
    – JohnFire
    May 2 '17 at 14:05
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    $\begingroup$ @JohnFire No, it means that $x^2$ has no inverse function on the real line. It does have an inverse function if considered on $[0, \infty)$, however. $\endgroup$
    – Arthur
    May 2 '17 at 14:08
  • $\begingroup$ Ah, okay @Arthur thank you :) $\endgroup$
    – JohnFire
    May 2 '17 at 14:10

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