## Calculus, 10th Edition (Anton)

$$f(x)=\sqrt{x+3}, \quad x \geq -3,$$ we find that: $$f^{-1}(x)=x^{2}-3 , \quad x \geq 0.$$
we first write $$y=f(x)=\sqrt{x+3}, \quad x \geq -3.$$ Then we solve this equation for $x$ as a function of $y$ $$y^{2}= x+3$$ $$x =y^{2}-3$$ which tells us that $$f^{-1}(y)=y^{2}-3. \quad \quad (i)$$ Since we want $x$ to be the independent variable, we reverse $x$ and $y$ in (i) to produce the formula $$f^{-1}(x)=x^{2}-3.$$ We know the domain of $f^{ −1}$ is the range of $f$ whereas the range of $f(x)=\sqrt{x+3}$ is $[0, \infty)$. Thus, if we want to make the domain of $f^{ −1}$ clear, we must express it explicitly by rewriting (i) as $$f^{-1}(x)=x^{2}-3 , \quad x \geq 0.$$