Question

Find the domain and the range of the real function $f\left(x\right)=\sqrt{9-x^2}$.

Answer 1

Domain and range

$f\left(x\right)=\sqrt{9-x^2}$

$y=\sqrt{9-x^2}$

$\sqrt{9-x^2}\ge 0$

$9-x^2\ge 0$

$x^2\le 9$

$x{\le }_{-}^{+}3$

$x\in [-3,3]$

$y^2=9-x^2$

$x^2=9-y^2$

$x=\sqrt{9-y^2}$

$9-y^2\ge 0$

$y^2\le 9y\in 3,-3$

$y=+ve$

$\because y\in [0,3]$