Answer
$$f^{-1}(x)=\frac{1}{\sqrt x}$$
- Domain: $(0,\infty)$.
- Range: $(0,\infty)$.
Work Step by Step
$$y=f(x)=\frac{1}{x^2}\hspace{1cm}x\gt0$$
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, $f(f^{-1}(x))=f^{-1}(f(x))=x$
