**Question:**

Find the domain and range of the following real function:

(i) $f(x)=-|x|$

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

**Solution:**

(i) $f(x)=-|x|, x \in \mathbf{R}$

We know that $|x|=\left\{\begin{array}{l}x, x \geq 0 \\ -x, x<0\end{array}\right.$

$\therefore f(x)=-|x|=\left\{\begin{array}{l}-x, x \geq 0 \\ x, x<0\end{array}\right.$

Since $f(x)$ is defined for $x \in \mathbf{R}$, the domain of $f$ is $\mathbf{R}$.

It can be observed that the range of $f(x)=-|x|$ is all real numbers except positive real numbers.

$\therefore$ The range of $f$ is $(-\infty, 0]$.

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

Since $\sqrt{9-x^{2}}$ is defined for all real numbers that are greater than or equal to $-3$ and less than or equal to 3, the domain of $f(x)$ is $\{x:-3 \leq x \leq 3\}$ or $[-3,3]$.

For any value of $x$ such that $-3 \leq x \leq 3$, the value of $f(x)$ will lie between 0 and 3 .

$\therefore$ The range of $f(x)$ is $\{x: 0 \leq x \leq 3\}$ or $[0,3]$.