What is the domain and range of this function and its inverse f(x) = sqrt(x + 7)?

Apr 19, 2015

Domain of f(x)= {x$\in$R, $x \ge - 7$}, Range= {y$\in$R, y$\ge 0$}
Domain of ${f}^{-} 1 \left(x\right)$= { x$\in$R}, Range = {y$\in$R, , $y \ge - 7$}

The domain of the function would be all x, such that $x + 7 \ge 0$, or $x \ge - 7$. Hence it is {x$\in$ R, $x \ge - 7$}

For range, consider y=$\sqrt{x + 7}$. Since$\sqrt{x + 7}$ has to be $\ge 0$, it is obvious that $y \ge 0$. Range would be {y$\in$R, y$\ge 0$}

The inverse function would be ${f}^{-} 1 \left(x\right)$= ${x}^{2} - 7$.

The domain of the inverse function is all real x that is { x$\in$R}

For the range of the inverse function solve y= ${x}^{2}$-7 for x. It would be x= $\sqrt{y + 7}$. This clearly shows that $y + 7 \ge 0$. Hence Range would be {y $\in$R, $y \ge - 7$}