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I have trouble with the following problem:

Prove that the function $f(x)=x^2-2x+3$, with domain $x\in (-\infty, 0)$, is a bijection from $(-\infty, 0)$ to its range.

Work: I tried to first prove one to one using proof by contrapositive. But I always get stuck and I cannot arrive at a conclusion like $a=b$. I think it maybe that I misinterpreted the question and I really need help.

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2 Answers 2

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$$f(x)=x^2-2x+3$$ For each $x\in(-\infty,0)$, there exists a unique $f(x)$, as is obvious, and no $x\in(-\infty,0)$ is mapped to more that one $f(x)$. Each value of $f(x)$ is paired with (i.e., paired as in $(x,f(x))$) one and one only $x\in(-\infty,0)$. Use the definition of a bijection.

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Hint: To show the function is $1-1$, prove that the function is strictly decreasing on the interval $(-\infty,0)$.

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