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I am not sure where to start or how to find a solution. How can I determine this is a one to one function:

$f(x) = x + \frac1{x - 1}$, for all real numbers $x \ne 1$.

Math_Ed_Student
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    With $x+1/x-1$ the restriction $x\ne 1$ does not make sense, so I assume that $x+\frac1{x-1}$ is what you meant. – Hagen von Eitzen Apr 12 '13 at 16:03
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3 Answers3

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Hint: Note that $f(x)=f(\frac1{x-1}+1)$.

Hagen von Eitzen
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We can calculate. Let $y=x+\frac{1}{x-1}$. For $x\ne 1$, this equation is equivalent to $x^2-(1+y)x+(1+y)=0$.

For a specific $y$, the quadratic equation has more than one real root $x$ precisely if the discriminant $(1+y)^2 -4(y+1)^2$ is $\gt 0$

There are clearly many $y$ for which this is true. For the discriminant simplifies to $(y+1)(y-3)$. So the function is very much not one to one. We can use the discriminant to determine precisely for which $y$ there are two solutions to the equation. These are $y\gt 3$ and $y\lt -1$.

Another way: We can be much more geometric. The resulting argument is nformal, but persuasive. (It can be made formal by using tools from the calculus.)

Note that $x+\frac{1}{x-1}$ blows up as $x$ approaches $1$ from the right. At say $x=2$, our function is quite reasonable, we have $f(2)=2$. And for large positive $x$ our function is large. So on $(1,\infty)$ our function goes smoothly from large positive vales, to modest values, and then to large positive values. So it must take the same value at least twice, indeed infinitely many times.

André Nicolas
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A different hint: Consider the definition of a one-to-one function:

If $f(a)=f(b) \implies a=b$, then $f$ is one-to-one.

clintonmonk
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