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Miscellaneous Exercise · Q3

Q.Find the domain of the function f(x)=x2+2x+1x2−8x+12f(x) = \dfrac{x^2 + 2x + 1}{x^2 - 8x + 12}.

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The domain of a rational function excludes any xx that makes the denominator zero. Here, solving x2−8x+12=0x^2 - 8x + 12 = 0 gives x=2x = 2 and x=6x = 6, so the domain is all real numbers except 22 and 66.

Why domain matters for rational functions

A rational function is a fraction of two polynomials. The only thing that can go wrong — the only place where the function is undefined — is when the denominator equals zero. Division by zero is not allowed in real numbers, so we must find every xx that makes the denominator zero and remove those values from the domain.

The numerator can be anything; it doesn't affect the domain at all. Even if the numerator is also zero at the same xx, the function is still undefined (that would give a 0/00/0 form, which is indeterminate, not a real number).

So the task reduces to: find all real zeros of the denominator, then state that the domain is R\mathbb{R} minus those points.

Step-by-step

  1. Identify the denominator. The denominator of f(x)f(x) is x2−8x+12x^2 - 8x + 12. We need to solve:

x2−8x+12=0x^2 - 8x + 12 = 0

  1. Factor the quadratic. Look for two numbers that multiply to +12+12 and add to −8-8. Those numbers are −2-2 and −6-6, because:

(−2)×(−6)=12and(−2)+(−6)=−8(-2) \times (-6) = 12 \quad\text{and}\quad (-2) + (-6) = -8

So the factorization is:

x2−8x+12=(x−2)(x−6)x^2 - 8x + 12 = (x - 2)(x - 6)

  1. Set each factor to zero.

x−2=0⇒x=2x - 2 = 0 \quad\Rightarrow\quad x = 2

x−6=0⇒x=6x - 6 = 0 \quad\Rightarrow\quad x = 6

  1. State the domain. The function is defined for every real xx except 22 and 66. In set notation:

Domain={x∈R∣x≠2 and x≠6}\text{Domain} = \{x \in \mathbb{R} \mid x \neq 2 \text{ and } x \neq 6\}

Or in interval notation:

(−∞,2)∪(2,6)∪(6,∞)(-\infty, 2) \cup (2, 6) \cup (6, \infty)

Watch out

A common mistake is to also exclude values that make the numerator zero. That is not correct — the numerator can be zero safely (the function value is just 00). Only the denominator matters for domain.

Tip

If the denominator had been something like x2+1x^2 + 1, which has no real zeros, the domain would be all real numbers. Always check the discriminant b2−4acb^2 - 4ac first: if it's negative, the denominator never hits zero, and the domain is R\mathbb{R}.

✓Final answer

The domain is all real numbers except x=2x = 2 and x=6x = 6: (−∞,2)∪(2,6)∪(6,∞)\boxed{(-\infty, 2) \cup (2, 6) \cup (6, \infty)}.

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