Let $f:(0,1)\to \mathbb R$ be defined by $f(x)=\frac{b−x}{1−bx}$, where $b$ is a constant such that $0<b<1$.
How do I prove that $f(x)$ is not invertible in $(0,1)$?
P.S:I was able to prove that $f(x)$ is one-one...
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If it is 1-1 then you must show it is not onto. Did you try graphing it for various $b$? – Gregory Grant Dec 31 '15 at 14:41
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Yeah I did.I know its not onto.But I'm not able to show it manually! @GregoryGrant – Dec 31 '15 at 14:42
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I did'nt get that "enough to show that" part @Hetebrij – Dec 31 '15 at 14:48
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Am I mistaken to think that since $f$ is injective then $f$ is invertible? – Jimmy R. Dec 31 '15 at 14:49
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@Stef Depends on the range. – Hetebrij Dec 31 '15 at 14:49
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@Hetebrij The range is $[-1,b]$ because $f$ is strictly decreasing. I do not see a problem. – Jimmy R. Dec 31 '15 at 14:52
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The range is stated to be $\Bbb R$. "Let $f:(0,1)\rightarrow\Bbb R$..." – Gregory Grant Dec 31 '15 at 14:56
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@Stef The range of $f$ is $\mathbb{R}$ by definition of $f$. We can restrict the range to $(-1,b)$ to make $f$ surjective. – Hetebrij Dec 31 '15 at 14:57
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The co-domain is $\Bbb R$. The range of $f$, namely $f((0,1))$, is $(-1,b)$. And $f$ is a bijection of $(0,1)$ onto $(-1,b)$. Thus $f$ is invertible, with inverse function $f^{-1}:(-1,b)\to(0,1)$ given by $f^{-1}(y) = {b-y\over 1-yb}$. – John Dawkins Dec 31 '15 at 15:44
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Let $a\in\Bbb R$ and suppose $f(x)=a$.
Then $\frac{b-x}{1-bx}=a$.
Solving for $x$ we have $x=\frac{b-a}{1-ba}$
Thus if $a=\frac1b$ then there can be no such $x$ that maps to it.
Thus $f$ is not onto. Thus $f$ is not invertible.
Gregory Grant
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Could it be made a little more rigorous?For many other values of a too corresponding values of x may not exist....can it be shown somehow?...i like the answer btw – Dec 31 '15 at 15:04
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@MathJack Yeah you can say, for example, that if $a<\frac1b$ then necessarily $a<1$. Is that required for your exercise? – Gregory Grant Dec 31 '15 at 15:10
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Yes.Could you please explain your last statement ^.I cant understand how you got it.. – Dec 31 '15 at 15:16
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@MathJack Sure, suppose $a<\frac1b$. Then you also need $a$ to be such that $x=\frac{b-a}{1-ba}\in(0,1)$, so you need to solve $0<\frac{a-b}{1-ba}<1$ and if you do you get $a<1$. You need $a<\frac1b$ so you know $1-ba$ is positive so when you clear the denominator it doesn't change the direction of the inequalities. – Gregory Grant Dec 31 '15 at 17:33
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@MathJack Solving the left hand side of the inequality we get $b<a$. So you have $b<a<1$. Now there's also the case $a>\frac1b$. That implies $a<b$ and it's impossible to have $\frac1b<a<b$ since $\frac1b>b$ (since $b\in(0,1)$). Thus you can say definitively that $b<a<1$. In other words the image of $f$ is $(b,1)$. Another proof that it's not onto, since $b>1$. – Gregory Grant Dec 31 '15 at 17:45
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Hey!Thank you soooooo much for explaining with such clarity :-D....you can add it to your answer too if you wish :-) – Dec 31 '15 at 17:50
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We have $f: (0,1) \to \mathbb{R}, x \mapsto \frac{b-x}{1-bx}$.
Now, we note that $f$ is continuous and injective, so in particular we must have that $f$ is increasing or decreasing.
Then we have $\hat{f}(0) = \lim_{x \to 0} f(x)$ is well-defined on the extended reals. In case $f$ is invertible, we must have $\hat{f}(0) = \pm \infty$, else the range of $f$ is bounded below or above, so in particular not the whole of $\mathbb{R}$, we have
$$ \hat{f}(0) = \lim_{x \to 0} \frac{b-x}{1-bx} = \frac{b}{1} = b$$
as the both parts of the fraction are continuous and the denominator is bounded away from zero.
As $0 < b <1$, we have that $f$ is not-invertible when defined from $(0,1) \to \mathbb{R}$.
Hetebrij
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$f(x)=b\iff b-x=b-b^2x\iff -x=-b^2x\iff x=0.$
DanielWainfleet
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f can be extended to a continuous function with domain [0,1]. The continuous image of a closed bounded interval is a closed bounded interval so the image of (0,1) is bounded. – DanielWainfleet Dec 31 '15 at 20:37
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$f'(x) = (-(1-bx)+b(b-x))/(1-bx)^2 = (b^2-1)/(1-bx)^2<0$ since $1<b$ thus $f$ decreases strictly and $f:(0,1)\rightarrow f(0,1)$ is bijective. But remak that $f(0,1)$ is not contained in $(0,1)$ since $f(1)=-1$. So $f$ is bijective onto its image.
Tsemo Aristide
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