Answer
$$f^{-1}(x)=1-\sqrt{x+1}$$
- Domain: $[-1,\infty)$.
- Range: $(-\infty,1]$
Work Step by Step
$$y=f(x)=x^2-2x\hspace{1cm}x\le1$$
To find its inverse:
1) Solve for $x$ in terms of $y$:
$$y=x^2-2x$$ $$y=(x^2-2x+1)-1$$ $$y=(x-1)^2-1$$ $$(x-1)^2=y+1$$
- Since $x\le1$, it follows that $(x-1)\le0$, so we will go with the negative value of $(x-1)$, which means:
$$x-1=-\sqrt{y+1}$$ $$x=1-\sqrt{y+1}$$
2) Interchange $x$ and $y$:
$$y=1-\sqrt{x+1}$$
Therefore, $$f^{-1}(x)=1-\sqrt{x+1}$$
- Domain: $x$ is defined where $(x+1)\ge0$, or $x\ge-1$.
So the domain of $f^{-1}$ is $[-1,\infty)$.
- Range: $\sqrt{x+1}\ge0$ for $x\in[-1,\infty)$
Thus $-\sqrt{x+1}\le0$, and $1-\sqrt{x+1}\le1$
That means the range of $f^{-1}$ is $(-\infty,1]$
*Check:
$$f(f^{-1}(x))=(1-\sqrt{x+1})^2-2(1-\sqrt{x+1})=[1+(x+1)-2\sqrt{x+1}]-2+2\sqrt{x+1}=x+2-2=x$$
$$f^{-1}(f(x))=1-\sqrt{x^2-2x+1}=1-\sqrt{(x-1)^2}=1-|x-1|$$
And since $x\le1$, we have $x-1\le0$, so $|x-1|=1-x$
$$f^{-1}(f(x))=1-(1-x)=1-1+x=x$$
Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$
