overrated3245w

2022-09-01

Find the inverse of a piecewise function, namely:

$f(x)=\{\begin{array}{ll}x-1,& 0\le x<1\\ 2-x,& 1<x\le 2\end{array}$

For the first case

$f:y=x-1$ when $0\le x<1\to {f}^{-1}:x=y+1$ when $-1\le x<0$

But I find this because I only have to subtract 1 from the endpoints of the inequality. I don't have a clear intuition how to do this with 2−x

$f(x)=\{\begin{array}{ll}x-1,& 0\le x<1\\ 2-x,& 1<x\le 2\end{array}$

For the first case

$f:y=x-1$ when $0\le x<1\to {f}^{-1}:x=y+1$ when $-1\le x<0$

But I find this because I only have to subtract 1 from the endpoints of the inequality. I don't have a clear intuition how to do this with 2−x

lascosasdeali3v

Beginner2022-09-02Added 10 answers

Suppose that ${f}^{-1}$ there exists, then you have $f(x)=x-1$ when $0\u2a7dx<1$ then setting $y=f(x)$ we get $y=x-1$ when $0\u2a7dx<1$ then $x=y+1$ when $0\u2a7dy+1<1$. Hence ${f}^{-1}(x)=x+1$ when $-1\u2a7dx<0$.Also $f(x)=2-x$ when $0<x\u2a7d2$ then setting y=f(x) we get $y=2-x$ when $1<x\u2a7d2$ then $x=2-y$ when $1<2-y\u2a7d2$. Hence ${f}^{-1}(x)=2-x$ when $0\u2a7dx<1$

Therefore, the inverse of function f is given by

$$\overline{){\displaystyle {f}^{-1}:x\mapsto \{\begin{array}{l}x+1,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}-1\u2a7dx<0,\\ 2-x,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}0\u2a7dx<1\end{array}}}$$

Now check $({f}^{-1}\circ f)(x)=x$ for all $x\in \mathrm{D}\mathrm{o}\mathrm{m}({f}^{-1})$ and $(f\circ {f}^{-1})(x)=x$ for all $x\in \mathrm{D}\mathrm{o}\mathrm{m}(f)$

Therefore, the inverse of function f is given by

$$\overline{){\displaystyle {f}^{-1}:x\mapsto \{\begin{array}{l}x+1,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}-1\u2a7dx<0,\\ 2-x,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}0\u2a7dx<1\end{array}}}$$

Now check $({f}^{-1}\circ f)(x)=x$ for all $x\in \mathrm{D}\mathrm{o}\mathrm{m}({f}^{-1})$ and $(f\circ {f}^{-1})(x)=x$ for all $x\in \mathrm{D}\mathrm{o}\mathrm{m}(f)$

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