Problem 60 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Graph \(f(x)=\frac{3 x^{2}-6 x+9}{x^{2}+x-2} .\) Determine the values of \(x\) where the function is undefined.

Step-by-Step Solution

Answer

The function is undefined at \( x = -2 \) and \( x = 1 \).

Step 1: Identify the denominator

The function is undefined when the denominator is equal to zero. Identify the denominator of the function. For the given function, the denominator is: \( x^2 + x - 2 \)

Step 2: Set the denominator to zero

Set the denominator equal to zero to find the values of \( x \) where the function is undefined. \( x^2 + x - 2 = 0 \)

Step 3: Solve for x

Solve the quadratic equation \( x^2 + x - 2 = 0 \) by factoring. \( (x + 2)(x - 1) = 0 \) Therefore, the solutions are: \( x + 2 = 0 \) or \( x - 1 = 0 \)\( x = -2 \) or \( x = 1 \)

Key Concepts

undefined values

In mathematics, a function is said to be undefined at certain points if its expression weighs down to a division by zero. For rational functions, which are ratios of polynomials, these are the points where the denominator is zero.
To determine where a function is undefined, follow these steps:

• Identify the denominator of the rational function.
• Set the denominator equal to zero.
• Solve the resulting equation to find the values of \(x\) that make the denominator zero.

In our example, the rational function is \(f(x)=\frac{3 x^{2}-6 x+9}{x^{2}+x-2}\). The first step is to look at the denominator \(x^2 + x - 2\). By setting this equal to zero and finding solutions, we get \(x = -2\) and \(x = 1\). Hence, \(f(x)\) is undefined at \(x = -2\) and \(x = 1\).

quadratic equations

Quadratic equations are a form of polynomial equations where the highest exponent of the variable is 2. The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). Solving quadratic equations requires finding values of \(x\) that satisfy the equation.
There are several methods to solve quadratic equations, including the quadratic formula, completing the square, and factoring. In the provided exercise, we used factoring to solve the quadratic equation \(x^2 + x - 2 = 0\).
Factoring turns the quadratic equation into a product of binomials. In our case, it factors to \((x + 2)(x - 1) = 0\). Thus, the solutions are obtained by setting each binomial equal to zero: \(x + 2 = 0\) or \(x - 1 = 0\), giving \(x = -2\) and \(x = 1\).

factoring

Factoring is a method used to simplify expressions and solve equations by expressing a polynomial as a product of its factors. This can be particularly helpful for solving quadratic equations.
In the exercise, the quadratic equation \(x^2 + x - 2 = 0\) was solved by identifying its factors. The process involves:

• Finding two numbers that multiply to give the constant term (-2) and add to give the coefficient of the linear term (1).
• These numbers are +2 and -1.
• Expressing the quadratic as \((x + 2)(x - 1) = 0\).

Setting each factor equal to zero gives the solutions. Factoring simplifies complex equations and aids in finding values where functions are undefined or zero. Practice helps in identifying factor pairs quickly and accurately.