Q.Find the domain and range of the following real functions:

Concept: Domain and range of real functions — for (i) an absolute value function, for (ii) a square root function.

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

Domain: $|x|$ is defined for all real $x$, so domain is $\mathbb{R}$.

Range: $|x| \ge 0 \implies -|x| \le 0$. The maximum value is $0$ (at $x=0$), and it decreases without bound as $|x| \to \infty$. Hence range is $(-\infty, 0]$.

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

Domain: require $9 - x^2 \ge 0 \implies x^2 \le 9 \implies -3 \le x \le 3$.

Range: $9 - x^2$ takes values from $0$ to $9$, so $\sqrt{9 - x^2}$ takes values from $0$ to $3$. Hence range is $[0, 3]$.

Concept First: What Domain and Range Really Mean

A real function takes a real number as input and gives a real number as output. The domain is the set of all real numbers you can safely plug in — numbers that don't break the function (like dividing by zero or taking the square root of a negative). The range is the set of all outputs the function actually produces as the input runs over the domain.

For $f(x) = -|x|$, there's no division or square root, so the only restriction is that $|x|$ is defined for every real $x$. That means the domain is all real numbers. The range? The absolute value is always non-negative, so $-|x|$ is always non-positive — zero or negative. The largest output is $0$ (when $x = 0$), and it goes down without bound as $|x|$ grows.

For $f(x) = \sqrt{9 - x^2}$, the square root demands that the inside be non-negative: $9 - x^2 \geq 0$. That gives $x^2 \leq 9$, so $x$ is between $-3$ and $3$. The range comes from the fact that the square root outputs only non-negative numbers, and the expression inside runs from $0$ to $9$, so the square root runs from $0$ to $3$.

Now let's work through each function carefully.

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

1. Domain

The absolute value function $|x|$ is defined for every real number $x$. There is no division, no square root, no other restriction. So the domain is simply all real numbers:

2. Range — the intuition

$|x|$ is always $\geq 0$. Multiplying by $-1$ flips the sign: $-|x|$ is always $\leq 0$. The smallest possible value of $|x|$ is $0$ (when $x = 0$), giving $f(0) = 0$. As $|x|$ grows without bound, $-|x|$ goes to $-\infty$. So the outputs cover all numbers from $-\infty$ up to and including $0$.

3. Formal range

Since $|x|$ takes every non-negative real value (as $x$ varies over $\mathbb{R}$), $-|x|$ takes every non-positive real value. Therefore:

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

1. Domain — the square root condition

The expression under the square root must be non-negative:

This rearranges to $x^2 \leq 9$, which means $|x| \leq 3$. So:

2. Range — what outputs are possible

Inside the square root, $9 - x^2$ runs from $0$ (when $x = \pm 3$) up to $9$ (when $x = 0$). The square root function $\sqrt{t}$ is increasing for $t \geq 0$, so:

• When $9 - x^2 = 0$, $\sqrt{0} = 0$.
• When $9 - x^2 = 9$, $\sqrt{9} = 3$.

Since $9 - x^2$ takes every value between $0$ and $9$ as $x$ varies over $[-3, 3]$, the square root takes every value between $0$ and $3$. Therefore: