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

$$f^{-1}(x)=\frac{1}{\sqrt x}$$ - Domain: $(0,\infty)$. - Range: $(0,\infty)$.
$$y=f(x)=\frac{1}{x^2}\hspace{1cm}x\gt0$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=\frac{1}{x^2}$$ $$x^2=\frac{1}{y}$$ $$|x|=\frac{1}{\sqrt y}$$ - Since $x\gt0$, it follows that $|x| = x$, which means: $$x=\frac{1}{\sqrt y}$$ 2) Interchange $x$ and $y$: $$y=\frac{1}{\sqrt x}$$ Therefore, $$f^{-1}(x)=\frac{1}{\sqrt x}$$ - Domain: $x$ is defined where $x\ge0$ and $x\ne0$, so it must be that $x\gt0$. So the domain of $f^{-1}$ is $(0,\infty)$. - Range: $\sqrt{x}\gt0$ for $x\in(0,\infty)$ Thus $\frac{1}{\sqrt x}\gt0$ and $\frac{1}{\sqrt x}$ reaches $\infty$. That means the range of $f^{-1}$ is $(0,\infty)$ *Check: $$f(f^{-1}(x))=\frac{1}{\Big(\frac{1}{\sqrt x}\Big)^2}=\frac{1}{\frac{1}{x}}=x$$ $$f^{-1}(f(x))=\frac{1}{\sqrt{\frac{1}{x^2}}}=\frac{1}{\frac{1}{|x|}}$$ And since $x\gt0$, we have $|x|=x$ $$f^{-1}(f(x))=\frac{1}{\frac{1}{x}}=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$