Q.Let $D$ be the domain of the real valued function $f$ defined by $f(x) = \sqrt{25 - x^2}$. Then, write $D$.

The key idea is that the expression under a square root must be non‑negative for the function to be real‑valued.

1. For $f(x) = \sqrt{25 - x^2}$ to be defined, we require $25 - x^2 \ge 0$.
2. This inequality rearranges to $x^2 \le 25$, which means $|x| \le 5$.
3. Therefore, $x$ lies in the closed interval $[-5, 5]$.

The core idea here is simple but crucial: the square root of a real number is only defined when the number inside is non-negative. You cannot take the square root of a negative number and get a real result — the function would become undefined in the real numbers. So the domain of $f$ is exactly the set of $x$ values that keep $25 - x^2$ at zero or above.

Let’s walk through it.

1. Set up the inequality.

   We need $25 - x^2 \ge 0$. This is the condition for $f(x)$ to be a real number.

2. Rearrange to a familiar form.

   $25 - x^2 \ge 0$ is equivalent to $x^2 \le 25$.

3. Solve the inequality $x^2 \le 25$.

   For any real $x$, $x^2 \le 25$ means that $x$ lies between $-5$ and $5$, inclusive. Why? Because if $x$ is greater than $5$ or less than $-5$, its square exceeds $25$.

   So the solution is $-5 \le x \le 5$.

4. Write the domain in interval notation.

   The set of all $x$ satisfying this is $[-5, 5]$.

Method: Domain of a square-root function

To find the domain of any $\sqrt{\text{expression}}$, require the expression under the root to be non-negative, then solve.

Steps

Step 1: Set the radicand $\ge 0$

The square root is real only when the inside is non-negative:

Step 2: Rearrange into a standard inequality

Step 3: Write the solution as an interval

$|x|\le 5$ means $-5\le x\le 5$, i.e. the domain is the closed interval $[-5,5]$.

For a root, use $\ge$ (equality allowed, since $\sqrt0=0$); reserve strict $>$ for expressions that also sit in a denominator.

Common Mistakes

Mistake 1: Keeping only the positive half, writing $x\le 5$

Why it's wrong: $x^2\le 25$ also allows negatives like $x=-4$ (since $16\le 25$). Correct approach: solve $x^2\le25$ as $-5\le x\le5$, both sides included.

Mistake 2: Using a strict inequality and excluding the endpoints

Why it's wrong: at $x=\pm5$ the radicand is $0$ and $\sqrt0=0$ is defined, so $\pm5$ belong to the domain. Correct approach: use $\ge$, giving the closed interval $[-5,5]$.

Mistake 3: Solving $25-x^2\ge0$ with a sign slip

Why it's wrong: moving $x^2$ across incorrectly can flip the inequality. Correct approach: rewrite as $x^2\le25$ carefully, then take square roots to $|x|\le5$.