## University Calculus: Early Transcendentals (3rd Edition)

$$f^{-1}(x)=1-\sqrt{x+1}$$ - Domain: $[-1,\infty)$. - Range: $(-\infty,1]$
$$y=f(x)=x^2-2x\hspace{1cm}x\le1$$ 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, $f(f^{-1}(x))=f^{-1}(f(x))=x$