Question

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

Answer 1

$\left(i\right)$ Given:$f\left(x\right)=-|x|,x\in R$
Step $1:$ Define function as we know, $|x|=\{\begin{array}{cc}x,& x\ge 0\\ -x,& x<0\end{array}$
$\therefore f\left(x\right)=-|x|=-\{\begin{array}{cc}x,& x\ge 0\\ -x,& x<0\end{array}$

$=\{\begin{array}{cc}-x,& x\ge 0\\ x,& x<0\end{array}$

Step $2:$ Domain of function
As $f\left(x\right)$ is defined for $x\in R$,the domain of $f$ is $R$.

Step $3:$ Draw diagram

Step $4:$ Range of function
We can see from graph, the range of $f\left(x\right)=-|x|$ is all real numbers except positive real numbers.
Therefore, the range of $f\left(x\right)$ is $(-\infty ,0]$.

$\left(ii\right)$ Given:$f\left(x\right)=\sqrt{9-x^2}$
Step $1:$ Domain of function
As we know , the expression inside root must be greater than or equal to zero.
So, $9-x^2\ge 0$
$\Rightarrow -(x^2-9)\ge 0$
Multiply both side by negative sign
$\Rightarrow (x^2-9)\le 0$
$\Rightarrow (x-3)(x+3)\le 0$
Domain of $f$ is $x\in [-3,3]$

Step $2:$ Range of function
As we know $x\in [-3,3]$ and we also know that minimum value of $x^2$ is zero.
Now,
Put $x=0,3$ and $-3$ function for finding range, we get
$f\left(0\right)=\sqrt{9-0}=3$
$f\left(3\right)=\sqrt{9-9}=0$
$f(-3)=\sqrt{9-9}=0$

From these values we can say $f\left(x\right)$ will lie betweemn $0$ and $3$
Therefore, the range of $f\left(x\right)$ is $[0,3]$