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Chapter 2: Problem 32

Find the domain of the function. $$ f(x)=\frac{1}{x^{2}+x-2} $$

Short Answer

Expert verified
The domain of the function \(f(x) = \frac{1}{x^{2}+x-2}\) is all real numbers x, except for \(x = -2\) and \(x = 1\). In other words, \(\text{Domain}(f(x)) = \{ x \in \mathbb{R} | x \neq -2,1 \}\).

Step by step solution

01

Identify the denominator of the function

In this case, the denominator of the function is given by \(x^{2}+x-2\).
02

Find the x-values that make the denominator equal to zero

To do this, we will solve the equation \(x^{2}+x-2 = 0\) for x. We can factor this quadratic equation as follows: \((x+2)(x-1) = 0\) Now, we can find the two x-values that satisfy the equation: \(x+2 = 0\) gives \(x = -2\) \(x-1 = 0\) gives \(x = 1\)
03

Remove the x-values that make the denominator zero from the domain

As we found in step 2, the denominator is equal to zero when \(x=-2\) or \(x=1\). Therefore, we must exclude these values from the domain of the function. The domain of the function is defined as: $$ \text{Domain}(f(x)) = \{ x \in \mathbb{R} | x \neq -2,1 \} $$ This means that the domain of the function \(f(x) = \frac{1}{x^{2}+x-2}\) is all real numbers, except for \(x = -2\) and \(x = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
Quadratic equations are polynomial equations of the second degree. They have the general form \(ax^2 + bx + c = 0\) where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). Solving quadratic equations is a fundamental skill in understanding higher algebra. There are several methods to solve quadratic equations, including:
  • Factoring: This involves expressing the quadratic equation as a product of two binomials. It is the most straightforward method when applicable.
  • Quadratic formula: Given as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), this formula provides the roots directly.
  • Completing the square: This method converts the equation into a perfect square trinomial, solving for \(x\).
The roots of the equation represent the values of \(x\) where the quadratic expression equals zero. In the context of our exercise, solving the quadratic equation \(x^2 + x - 2 = 0\) helped identify where the denominator becomes zero.
Factoring Polynomials
Factoring polynomials is an essential technique to simplify algebraic expressions and solve equations. It involves breaking down a polynomial into a product of simpler polynomials. For quadratic expressions, like \(x^2 + x - 2\) in our exercise, the process typically involves:
  • Looking for two numbers: These numbers should multiply to the constant term (in this case, \(-2\)) and add up to the coefficient of the linear term (here \(1\)).
  • Simplifying into binomials: For \(x^2 + x - 2\), factoring gives \((x + 2)(x - 1)\).
Factoring allows us to easily find the zeros of the expression, which are crucial for determining where the function \(f(x) = \frac{1}{x^2 + x - 2}\) is undefined, specifically the \(x\)-values that make the denominator zero.
Rational Functions
Rational functions are expressions that involve ratios of polynomials. They take the form \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x) eq 0\) to avoid division by zero. Determining the domain of rational functions is about identifying all possible input values \(x\) where the function is defined:
  • Undefined points: These are determined by finding the \(x\)-values that make the denominator zero, as seen in our exercise with \(x = -2\) and \(x = 1\).
  • Domain specification: Once these points are known, the domain is expressed as all real numbers except these problematic \(x\)-values, i.e., \( \{ x \in \mathbb{R} | x eq -2, 1 \} \).
Understanding rational functions involves not only operations with polynomials but also recognizing their restrictions to describe their valid input scope, which is crucial in many areas of mathematics and applied sciences.

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Most popular questions from this chapter

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