Question

Find the domain and range of the following real functions:
(i) $f\left(x\right)=-\mid x\mid$ (ii) $f\left(x\right)=\sqrt{9-x^2}$

Answer 1

As we are given a real function, thus the domain and range should be real numbers.

(i) $f\left(x\right)=-\left|x\right|$

$f(-1)=-|-1|=-1$

$f\left(0\right)=-\left|0\right|=0$

$f\left(1\right)=-\left|1\right|=-1$

Here, $x$ can be any real number, but $f\left(x\right)$ will always be negative or zero.

Therefore,

Domain of function $=R\left(\text{All real numbers}\right)$

Range of the function $=$ Negative real numbers

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

$9-x^2\ge 0$

$x^2<9$

$x<\pm 3$

$x\in [-3,3]$

$f(-3)=\sqrt{9-{(-3)}^2}=0$ (Real number)

$f\left(0\right)=\sqrt{9-0}=3$ (Real number)

$f\left(3\right)=\sqrt{9-\left(3\right)}=0$ (Real number)

Here,

$-3\le x\le 3$

$0\le f\left(x\right)\le 3$

Therefore,

Domain of function $=[-3,3]$

Range of the function $=[0,3]$