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Let there be a function $f: \mathbb R \to \mathbb R$ given by $$f(x)=\begin{cases}5x + 2&x\geq 1\\x-1&x<1 \end{cases}$$ Give an example of a Left inverse of $f$, and prove that it is correct.

I can do this with normal function. But how does this work when we try a piece defined function?

AlexR
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arieenhenk
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    First, you do it on each piece separately. Then you have to work out the range of the argument that maps to each piece. Since $f(x)$ is not onto $\Bbb R$, there will be some arguments of the inverse that can be anything. – Ross Millikan Oct 07 '13 at 17:30

1 Answers1

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Jasper Loy's answer here already shows you most of what you need in order to define $g:\Bbb R\to\Bbb R$ such that $g\circ f=id_{\Bbb R}$. It doesn't actually matter what $g(x)$ is for $0\le x<7,$ since $f$ never takes on values in that interval. Thus, we can define $g$ by $$g(x)=\begin{cases}\frac{x-2}5 & \text{for }x\ge 7\\x+1 & \text{for }x<0\\\text{just about any real-valued function you want} & \text{for }0\le x<7.\end{cases}$$

You can also take a look at this answer to see how one should proceed to construct a left inverse in general.

Community
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Cameron Buie
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