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

$$f^{-1}(x)=b-\sqrt{x+b^2}$$ - Domain: $[-b^2,\infty)$. - Range: $(-\infty, b]$
$$y=f(x)=x^2-2bx\hspace{1cm}b\gt0\hspace{1cm}x\le b$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=x^2-2bx$$ $$y=(x^2-2bx+b^2)-b^2$$ $$y=(x-b)^2-b^2$$ $$(x-b)^2=y+b^2$$ $$|x-b|=\sqrt{y+b^2}$$ As $x\le b$, we take $|x-b|=b-x$ $$b-x=\sqrt{y+b^2}$$ $$x=b-\sqrt{y+b^2}$$ 2) Interchange $x$ and $y$: $$y=b-\sqrt{x+b^2}$$ Therefore, $$f^{-1}(x)=b-\sqrt{x+b^2}$$ - Domain: $x$ is defined where $(x+b^2)\ge0$, or $x\ge -b^2$. So the domain of $f^{-1}$ is $[-b^2,\infty)$. - Range: We have $\sqrt{x+b^2}\ge0$ Therefore, $-\sqrt{x+b^2}\le0$ And $b-\sqrt{x+b^2}\le b$ That means the range of $f^{-1}$ is $(-\infty, b]$ *Check: $$f(f^{-1}(x))=(b-\sqrt{x+b^2})^2-2b(b-\sqrt{x+b^2})=b^2-2b\sqrt{x+b^2}+(x+b^2)-2b^2+2b\sqrt{x+b^2}=2b^2+x-2b^2=x$$ $$f^{-1}(f(x))=b-\sqrt{x^2-2bx+b^2}=b-\sqrt{(x-b)^2}=b-|x-b|$$ We show above that $|x-b|=b-x$: $$f^{-1}(f(x))=b-(b-x)=b-b+x=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$