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I have a function

$h\left(x\right)=\sqrt{\mu x + \eta}$ such that $\mu\in\left(0,+\infty\right)$ and $\eta\in\mathbb{R}\setminus\left\{0\right\}$.

I want to check the domain and range of $h$ in case the coefficient $\eta<0$ or $\eta>0$.

To do so, I proceeded as$\colon$

$h\left(x\right)=\begin{cases} \sqrt{\mu x + \eta}, \hspace{20mm} \eta>0\\ \sqrt{\mu x - \eta}, \hspace{20mm} \eta<0\\ \end{cases} $

and concluded that the domain of $h$, namely $D_h$ is found as

$D_h=\begin{cases} x\in\left[\frac{-\eta}{\mu},+\infty\right), \hspace{20mm} \text{since}\hspace{2mm}\sqrt{\mu x + \eta}>0 \\ x\in\left[\frac{+\eta}{\mu},+\infty\right), \hspace{20mm} \text{since}\hspace{2mm}\sqrt{\mu x - \eta}>0 \end{cases} $

Also, I found the range of $h$, namely $R_h=\left[0,+\infty\right)$ in both above cases.

Now, I want to observe that under what conditions $h$ is a contraction mapping, namely $R_h\subset D_h$.

In case $\eta<0$, then I need to have

$\underbrace{\left[0,\infty\right)}_{R_h} \subset \underbrace{\left[\frac{+\eta}{\mu},+\infty\right)}_{D_h} \implies \frac{\eta}{\mu}<0 \hspace{5mm} \implies \hspace{5mm} \eta<0 \hspace{5mm} \text{But I already have this satisfied.} $

And In case $\eta>0$, then I need to have

$\underbrace{\left[0,\infty\right)}_{R_h}\subset \underbrace{\left[\frac{-\eta}{\mu},+\infty\right)}_{D_h} \implies \frac{-\eta}{\mu}<0 \hspace{5mm} \implies \hspace{5mm} \eta>0 \hspace{5mm} \text{But I already have this satisfied.} $

Is this approach correct?

user572171
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  • The sign on $\eta$ shouldn't change under the radical based on its positivity or negativity. The domain for $h$ should be computed just by considering the inequality $\mu x + \eta \geq 0$. Considerations about what this domain really "looks like" (e.g. whether it includes 0 or not) should only be made after finding a lower bound for $x$ which depends on $\mu$ and $\eta$, and at this stage you should consider whether $\eta$ is positive or negative. – TM Gallagher Feb 22 '19 at 14:31
  • Thanks. Fully, understood. A question : Why can not I originally write my function as $h\left(x\right)=\sqrt{\mu x - \eta}$ instead of $h\left(x\right)=\sqrt{\mu x + \eta}$ and then discuss about the sign of the $\eta$? – user572171 Feb 22 '19 at 14:43
  • Of course you can do that, and you will get the exact same conclusions from the same analysis if you write it as $-\eta$ instead of $\eta$. The key is consistency. If you write it as $\sqrt{\mu x + \eta}$ in the original definition of your function, you should use that format in your analysis that follows. It was in switching from $+\eta$ to $-\eta$ in the middle of the analysis that made things a little strange. – TM Gallagher Feb 22 '19 at 19:20
  • These are great advises. Thanks! Meanwhile, I could come out with a totally new proof. It looks much more convenient. I still work on it and would be sharing it soon :) – user572171 Feb 25 '19 at 09:13

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