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Let $f: (\mathbb{R},|.|)\to (\mathbb{R},|.|)$ defined by $f(x)=\dfrac{1}{1+x^2}$

I want to find $f^{-1}(]a,b[)$

What I do: $$f^{-1}(]a,b[)=\{y\in \mathbb{R}, \frac{1}{1+y^2}\in ]a,b[\}$$

that is : $$a<\frac{1}{1+y^2}<b$$

  • If $a>0$ then $$\frac1b-1<y^2<\frac1a-1$$ after that we must suppose two sases $\frac1b\leq 1 $ or $\frac1b>1$

Is there an other method ?

i suppose that $a<b$ $$ f^{-1}(]a,b[)= \begin{cases} \emptyset,~ b<0\\ ]-\infty,-\sqrt{\frac1b-1}[\cup]\sqrt{\frac1b-1},+\infty[,~ a\leq0 ~\text{and}~b<1\\ \mathbb{R},~ b>1,a\leq0\\ \mathbb{R}\setminus{0},~ b=1,a\leq0\\ \emptyset,~b\geq1,a>1\\ ]-\sqrt{\frac1a-1}, \sqrt{\frac1a-1}[,~b\geq1,a<1\\ ]\sqrt{\frac1b-1},\sqrt{\frac1a-1}[,~ b<1, a>0 \end{cases} $$

Thank you

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    $\begingroup$ You are on the right track. You laso have to consider the case $a<0$ and $b>0$. $\endgroup$ Mar 1, 2019 at 10:24
  • $\begingroup$ There are three cases to consider $a > 1, a \leq 1 < b$ and $a < b \leq 1.$ $\endgroup$
    – little o
    Mar 1, 2019 at 10:28
  • $\begingroup$ @KaviRamaMurthy please see my edits and tel me if it is correct $\endgroup$
    – user523857
    Mar 1, 2019 at 10:56
  • $\begingroup$ @Dbchatto67 please see my edits and tel me if it is correct $\endgroup$
    – user523857
    Mar 1, 2019 at 10:57

1 Answer 1

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I think a good way to approach the problem is to think first about the function $f$:

First of all, since $x^{2} \geq 0$ then $1+x^{2} \geq 1$ which leads to $ f(x) = \frac{1}{1+x^{2}} \leq 1$. This means that if $a > 1$ then $f^{-1}(]a,b[) = \emptyset$.

We also have that $1+x^{2} \geq 1 > 0$ so $f(x) = \frac{1}{1+x^{2}} >0$. This means that if $b \leq 0$ then we also have $f^{-1}(]a,b[) = \emptyset$.

Furthermore, $\forall a,a' \in ]\infty, 0]$ and $b>0$ we have $f^{-1}(]a,b[) = f^{-1}(]a',b[)$ and $\forall a < 1$ and $b \in ]1, \infty]$ we have $f^{-1}(]a,b[) = f^{-1}(]a,b'[)$

With this results, we can then separate the problem in appropriate cases:

$\underline{\text{Case} \,\, 1:}$ $0<a<b \leq 1$

$$a < f(x) < b \,\,\Leftrightarrow \,\, a < \frac{1}{1+x^{2}} < b \,\,\overset{a,b >0}{\Longleftrightarrow}\,\, \frac{1}{b} < 1+x^{2} < \frac{1}{a} \,\,\Leftrightarrow \,\,\frac{1}{b}-1 < x^{2} < \frac{1}{a}-1$$

Now, since $a, b\leq 1$, we have that $\frac{1}{b}-1 \geq 0$ and $\frac{1}{a}-1 \geq 0$ and we can take square roots to have

$$ \frac{1}{b}-1 < x^{2} < \frac{1}{a}-1 \,\, \Leftrightarrow \,\, \sqrt{\frac{1}{b}-1} < |x| < \sqrt{\frac{1}{a}-1}$$

(notice that the fact of the square root being an increasing injective function allows us to take square roots and not changing the grater-than and less-than symbols and to have the equivalence symbol)

We finally have that $$ x \in f^{-1}(]a,b[) \,\,\Leftrightarrow \,\, x\in \,\, ]\sqrt{\frac{1}{b}-1},\sqrt{\frac{1}{a}-1}[ \quad \text{or} \quad x\in \,\, ]-\sqrt{\frac{1}{a}-1},-\sqrt{\frac{1}{b}-1}[$$

We put now our attention on the limit cases:

$\underline{\text{Case} \,\, 2:}$
$\underline{\text{Subcase} \,\, 1:}$ $a=0$ and $b \leq 1$

$a=0 < \frac{1}{1+x^{2}}$ so the condition $a<f(x)<b$ is equivalent to $f(x)<b$.

$\underline{\text{Subcase} \,\, 2:}$ $a>0$ and $b > 1$

We have $\frac{1}{1+x^{2}} \leq 1<b$ so the condition $a<f(x)<b$ is equivalent to $a<f(x)$.

We can summarize the problem in the next cases:

The empty cases:

$a > 1\, \Rightarrow \,f^{-1}(]a,b[) = \emptyset$

$b \leq 0 \, \Rightarrow \,f^{-1}(]a,b[) = \emptyset$

The whole case:

$a \leq 0$ and $b >1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = \mathbb{R}$

The partial cases:

$a \leq 0$ and $0 < b \leq 1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = ]-\infty,-\sqrt{\frac{1}{b}-1}[ \,\, \bigcup \,\, ]\sqrt{\frac{1}{b}-1},\infty[$

$1 > a > 0$ and $ b > 1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = ]-\sqrt{\frac{1}{a}-1},\sqrt{\frac{1}{a}-1}[ $

The restrictive case:

$0<a<b \leq 1$ and $0 < b \leq 1$ $ \, \Rightarrow \,f^{-1}(]a,b[) = ]-\sqrt{\frac{1}{a}-1},-\sqrt{\frac{1}{b}-1}[ \,\, \bigcup \,\, ]\sqrt{\frac{1}{b}-1},\sqrt{\frac{1}{a}-1}[$

Notice that the good choice of the limit points ($0$ for $a$ and $1$ for $b$) is the key.

The proof may appear very long but if you draw the function (maybe with WolframAlpha) you will understand it much better.

I hope it helps you!

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  • $\begingroup$ please if I want to prove that $]x-r,x+r[\subset f^{-1}(]x-\varepsilon, x+\varepsilon[)$ without find exactly the pre image how to do please? $\endgroup$
    – user523857
    Apr 13, 2019 at 19:46
  • $\begingroup$ This is not true in general... What do you mean exactly? prove the continuity of the function by definition? $\endgroup$
    – Manza
    Apr 14, 2019 at 20:36

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