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For a long time I used the Domain of the inverse of a function to find the original function’s range. That's a little strategy that I use. Recently I've come across an example in which that doesn't seem to apply.

The function I came across is

$$ f(x) = 1 + \sqrt{2+3x} $$

and it’s inverse

$$ f^{-1}(x) = \frac{(x-1)^2-2}{3} $$

If you just look at the inverse, you are tempted to say that the range of the original function $f$ is all real numbers, and that is not the case. We know that

$$ D_f: x \geq - \frac{2}{3} $$ $$ R_f: y \geq 1 $$

So my question is: when can I not use that strategy?

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    $\begingroup$ An inverse function must be injective on its domain. Your function isn't on IR. Another example is just to take the root of x. Its inverse is x^2, which can also be defined on IR - but not as the inverse. $\endgroup$ – Friedrich Philipp Mar 8 '16 at 14:35
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You can always use this strategy. The problem is that the function $$g(x)= \frac{(x-1)^2-2}{3}$$ is not an inverse of $f$ (not for all $x$). When calculating it you should proceed as follows:

$$y=1+\sqrt{2+3x}$$ $$y-1=\sqrt{2+3x}$$

But now $\sqrt\cdot$ is non-negative, so is $y-1$ so $y\geqslant 1$. Then

$$2+3x=(y-1)^2$$ $$x=\frac{(y-1)^2-2}{3}\quad\text{provided }y\geqslant 1$$

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  • $\begingroup$ I see what you mean, thank you for clarifying that for me. Cheers! $\endgroup$ – bru1987 Mar 8 '16 at 14:50

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