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Let $f:\mathbb{R}^+ \to \mathbb{R}^+$ be given by $$f(x)=\frac{x + a}{x + b}.$$ The constants $a$ and $b$ must be evaluated to the following:

  • $a$ is the sum of the first 4 digits of your student number
  • $b$ is the product of the last two non-zero digits of your student number.

Examples:

  • If your student number is $30247340$, then $a = 9$ and $b = 12$.
  • If your student number is $1002110833567$, then $a = 3$ and $b = 42$.

Determine if $f$ is a bijection. If it is a bijection, find its inverse.

I used contradiction to show that $f$ is one to one, but I have no idea how to prove it is not onto. Since I don't know the exact values of $a$ and $b$, I am unable to find a counter-example to show that it is not onto. What other approaches can I use?

Chris Uchizono
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betty12340
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    For what value of $x$ is $f(x) = 1$? – sudeep5221 Nov 25 '22 at 21:41
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    If $a=b$, then $f$ is not one-to-one. – Taladris Nov 25 '22 at 21:46
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    Let's hope your student number is not $5034000$ – Henry Nov 25 '22 at 21:46
  • "If your student number is 30247340, then a=9 and b=12 . If your student number is 1002110833567 , then a=3 and b=42." Then $a$ and $b$ aren't constant, are if they different from student to student. Is $f$ a different function for different students. As Henry points out if you number is such that $a = b$ then the function is the constant function $f(x) =1$. – fleablood Nov 25 '22 at 23:04
  • But in general the way to show if something is surgective is to set $f(x) = m$ for any $m\in \mathbb R^+$ and solve for $x$. If $x$ is well defined for all possible $m$ it is surjective. $f(x)=\frac {x+a}{x+b}= m\implies x+a=m(x+b)\implies mx-x = a-mb\implies x(m-1)=a-mb$. If $m\ne 1$ we have $x =\frac{a-mb}{m-1}$ will be so that $f(x) =m$. But if $m = 1$ we are skewered six ways to sunday. $f(x)=1$ is impossible to solve if $a\ne b$. And if $a=b$ then $(x)=\frac{x+a}{x+b} =1$ and $f(x) \ne 1$ is impossible to solve. Either way, it's not surjective. (And it's not injective if $a=b$). – fleablood Nov 25 '22 at 23:11

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