Find the domain and Range of the function $f(x) = \dfrac{1}{\sqrt{x - 5}}$.

Find the domain and Range of the function f(x) = d – Class 11 Mathematics

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f(x) = \dfrac{1}{\sqrt{x - 5}}

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Concept understanding — Rational Function Domain

What is a Rational Function Domain?

Imagine you're baking a cake and the recipe says "add flour until the mixture is smooth." If you add too much flour, the mixture becomes a dry lump — it stops being a proper batter. A rational function is like that mixture: it's a fraction made of two polynomials, and it only "works" when the denominator isn't zero.

A rational function looks like this:

f(x) = \frac{P(x)}{Q(x)}

where $P(x)$ and $Q(x)$ are polynomials, and $Q(x) \neq 0$ .

The domain of a rational function is simply the set of all real numbers $x$ for which the function is defined — meaning, all $x$ except those that make the denominator zero.

The Intuition First

Think of division in everyday life. You can divide 10 apples among 5 people — that's fine. You can divide 10 apples among 2 people — also fine. But can you divide 10 apples among 0 people? That doesn't make sense. You can't split something among nobody.

In the same way, a rational function is a division. The denominator tells you "how many groups" you're splitting into. If the denominator is zero, the division is impossible — the function has no value there.

So the domain is: all real numbers, except the ones that make the bottom zero.

The Precise Statement

\text{Domain of } f(x) = \frac{P(x)}{Q(x)} \text{ is } \{ x \in \mathbb{R} \mid Q(x) \neq 0 \}

In plain words: find every $x$ that makes $Q(x) = 0$ , and remove those from the set of all real numbers.

How to Find the Domain — Step by Step

Step 1: Write down the denominator $Q(x)$ .

Step 2: Set $Q(x) = 0$ and solve for $x$ .

Step 3: The domain is all real numbers except those solutions.

Tip

You only care about the denominator. The numerator $P(x)$ can be anything — even zero — and the function is still defined (it just equals zero). Only the denominator matters for domain.

Examples

Example 1: $f(x) = \frac{1}{x-3}$

Denominator: $x - 3 = 0 \implies x = 3$

Domain: all real numbers except $3$ . In interval notation: $(-\infty, 3) \cup (3, \infty)$

Example 2: $f(x) = \frac{x^2 + 1}{x^2 - 4}$

Denominator: $x^2 - 4 = 0 \implies (x-2)(x+2) = 0 \implies x = 2$ or $x = -2$

Domain: all real numbers except $2$ and $-2$ . In interval notation: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

Example 3: $f(x) = \frac{2x+5}{x^2 + 1}$

Denominator: $x^2 + 1 = 0 \implies x^2 = -1$ — no real solution.

Domain: all real numbers, i.e., $(-\infty, \infty)$ …

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The function $f(x) = \frac{1}{\sqrt{x - 5}}$ requires $x - 5 > 0$ (positive under the square root, non-zero in denominator), giving domain $(5, \infty)$ ; as $x$ ranges over this interval, $f(x)$ takes all positive values, so range is $(0, \infty)$ .

Understanding the constraints

A rational function with a square root in the denominator imposes two simultaneous conditions. The expression under the square root must be non-negative (since we're working with real numbers), and the denominator itself cannot be zero. These aren't separate checks that we combine—they work together to carve out the valid input values.

For $f(x) = \frac{1}{\sqrt{x - 5}}$ , the square root $\sqrt{x - 5}$ appears in the denominator. This immediately tells us that $x - 5$ must be strictly positive, not merely non-negative. If $x - 5 = 0$ , the square root would be zero and we'd be dividing by zero. If $x - 5 < 0$ , the square root wouldn't be real.

Finding the domain

Set up the inequality. We need $x - 5 > 0$ (strict inequality, not $\geq$ ).
Solve for $x$ . Adding $5$ to both sides gives $x > 5$ .
Express in interval notation. The domain is all real numbers strictly greater than $5$ , which we write as $(5, \infty)$ .

Watch out

A common mistake is writing the domain as $[5, \infty)$ , including the endpoint. But $x = 5$ makes the denominator zero, so it must be excluded. The parenthesis matters.

Finding the range

Now we ask: as $x$ varies over $(5, \infty)$ , what values can $f(x)$ take?

Analyze the behavior near the boundary. As $x \to 5^+$ (approaching $5$ from the right), the quantity $x - 5 \to 0^+$ , so $\sqrt{x - 5} \to 0^+$ , and therefore $f(x) = \frac{1}{\sqrt{x - 5}} \to +\infty$ .
Analyze the behavior at infinity. As $x \to \infty$ , we have $x - 5 \to \infty$ , so $\sqrt{x - 5} \to \infty$ , and thus $f(x) = \frac{1}{\sqrt{x - 5}} \to 0^+$ . …

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Q.Find the domain and Range of the function f(x)=1x−5f(x) = \dfrac{1}{\sqrt{x - 5}}f(x)=x−5​1​.

Understanding the constraints

Finding the domain

Finding the range

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Q.Find the domain and Range of the function $f(x) = \dfrac{1}{\sqrt{x - 5}}$ .

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