Q.The domain of the function $f$ given by $f(x) = \dfrac{x^2 + 2x + 1}{x^2 - x - 6}$
(A) $\mathbf{R} - \{3,\ -2\}$
(B) $\mathbf{R} - \{-3,\ 2\}$
(C) $\mathbf{R} - [3,\ -2]$
(D) $\mathbf{R} - (3,\ -2)$

The key idea is that the domain of a rational function includes all real numbers for which its denominator is non-zero.

To find the domain of $f(x) = \dfrac{x^2 + 2x + 1}{x^2 - x - 6}$, we must ensure the denominator is not equal to zero.

Set the denominator to zero and solve for $x$:

$x^2 - x - 6 = 0$

Factor the quadratic expression:

$(x - 3)(x + 2) = 0$

This yields two values for $x$ that make the denominator zero:

$x - 3 = 0 \implies x = 3$ …

When we talk about the domain of a function, we are identifying all possible input values ($x$) for which the function produces a real, defined output ($f(x)$). For different types of functions, different restrictions apply.

For a rational function, which is a ratio of two polynomials, the primary restriction comes from the denominator. Division by zero is undefined in mathematics. Therefore, any value of $x$ that makes the denominator equal to zero must be excluded from the function's domain. The numerator, being a polynomial, is defined for all real numbers, so it does not introduce any restrictions on its own.

Let's find the domain of the given function step-by-step.

1. Identify the function type:

   The given function is $f(x) = \dfrac{x^2 + 2x + 1}{x^2 - x - 6}$. This is a rational function because it is expressed as a ratio of two polynomials: $P(x) = x^2 + 2x + 1$ (the numerator) and $Q(x) = x^2 - x - 6$ (the denominator).

2. State the condition for the domain:

   For a rational function, the domain consists of all real numbers for which the denominator is not equal to zero.

   So, we must have $x^2 - x - 6 \neq 0$.

3. Find the values of $x$ that make the denominator zero:

   To find the values that must be excluded, we set the denominator equal to zero and solve for $x$:

This is a quadratic equation. We can solve it by factoring. We need two numbers that multiply to $-6$ and add up to $-1$. These numbers are $-3$ and $2$.
So, we can factor the quadratic as:

For this product to be zero, at least one of the factors must be zero:

Solving these linear equations gives:

These are the values of $x$ for which the denominator becomes zero, and thus, the function $f(x)$ is undefined.

4. Express the domain:

The domain of $f(x)$ includes all real numbers except $3$ and $-2$. We express this using set notation. The set of all real numbers is denoted by $\mathbf{R}$. The set of values to be excluded is $\{3, -2\}$. …