Problem 11 - Precalculus Mathematics for Calculus

Question

Find the domain of the expression. $$\frac{x^{2}+1}{x^{2}-x-2}$$

Step-by-Step Solution

Answer

The domain is \(( -\infty, -1 ) \cup (-1, 2) \cup (2, \infty)\)."

Step 1: Understand the Problem

We need to find the domain of the expression \( \frac{x^2+1}{x^2-x-2} \). This means we need to determine all possible values of \( x \) for which the expression is defined.

Step 2: Identify Restrictions

The expression is a fraction, so the denominator cannot be zero. We must find the values of \( x \) for which the denominator \( x^2-x-2 \) equals zero, as these are the values that will make the expression undefined.

Step 3: Factor the Denominator

To find when \( x^2-x-2 = 0 \), we factor it into \( (x-2)(x+1) = 0 \).

Step 4: Solve the Equation

Set each factor equal to zero: \( x-2 = 0 \) gives \( x = 2 \) and \( x+1 = 0 \) gives \( x = -1 \). These are the values where the denominator is zero, so the expression is undefined.

Step 5: Write the Domain

The domain of the expression is all real numbers except \( x = 2 \) and \( x = -1 \). In interval notation, this is written as \(( -\infty, -1 ) \cup (-1, 2) \cup (2, \infty)\).

Key Concepts

Domain of a Function

The domain of a function refers to all possible values of the variable that makes the function work properly. For rational functions, like \( \frac{x^2+1}{x^2-x-2} \), we concentrate on when the denominator isn't zero. If the bottom of the fraction turns into zero, the function gets undefined.
Imagine you have a bridge, and you're looking for points that could break and fail. Similarly, the points that make the denominator zero cause 'breaks' in the function.
To find these, you'll need to determine when the denominator equals zero. For this function, these values are where the bottom part, \( x^2-x-2 \), doesn't work (equals zero, actually). Solving \( x^2-x-2=0 \) tells us where these issues happen.

Factoring Polynomials

Factoring polynomials is like finding what numbers multiply together to get the original number. It transforms the polynomial into simpler parts, which makes it easier to solve.
Let's see what's in action with our example: \( x^2-x-2 \). You're trying to break this down to find out when it hits zero.
In this case, you rewrite it as \( (x-2)(x+1) \). These two parts, when multiplied together, give back your original polynomial.
To factor correctly, you usually look for two numbers that multiply to give the constant term (in our case \(-2\)) and add up to the middle coefficient (\(-1\)).

• Here, \(2\) and \(-1\) multiply to \(-2\) and add to \(-1\), fitting perfectly.

Factoring is key because it directly points out the values that would turn the denominator into zero.

Interval Notation

Interval notation is a compact way of describing where a function is valid. It uses parentheses \(( \text{and} )\) and brackets \([ \text{and} ]\) to show where numbers start and stop.

If you have a number that makes a function undefined, exclude it by using parentheses \( ) \). In our exercise, we learned there are problems at \( x = -1 \) and \( x = 2 \).
In interval notation, this is written as \( ( -\infty, -1 ) \cup (-1, 2) \cup (2, \infty) \).

• \(( -\infty, -1 )\) means from negative infinity to just before \(-1\).
• \((-1, 2)\) covers everything between \(-1\) and \(2\), not including \(-1\) and \(2\).
• \((2, \infty)\) goes from just after \(2\) onwards.

Think of it as mapping out territories where hostility (undefined values) won't be met!