Problem 36 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

List all values of \(x\) for which the given function is not continuous. $$ g(x)=\frac{x^{3}+5 x}{(x-2)(2 x+3)} $$

Step-by-Step Solution

Answer

The function is not continuous at \(x = 2\) and \(x = -\frac{3}{2}\).

Step 1: Identify the Domain

First, find the values of x that make the denominator equal to zero, because these are the points where the function might be discontinuous. Set \[ (x-2)(2x+3) = 0 \] and solve for x.

Step 2: Solve for x in Factored Form

Solve each factor separately:1. \[ x - 2 = 0 \]2. \[ 2x + 3 = 0 \]This gives \[ x = 2 \] and \[ x = -\frac{3}{2} \].

Step 3: Verify Discontinuity Points

Check the values obtained. Both \(x = 2\) and \(x = -\frac{3}{2}\) make the denominator zero, thus leading to vertical asymptotes. Hence, the function is not continuous at these points.

Key Concepts

Domain of a Function

In mathematics, the domain of a function is the set of all possible input values (typically denoted as x) for which the function is defined. For a rational function like \( g(x) = \frac{x^{3} + 5x}{(x-2)(2x+3)} \), the domain is all real numbers except where the denominator equals zero. This is because a zero in the denominator makes the function undefined, leading to potential discontinuities.

To find these points:

• We set the denominator \( (x-2)(2x+3) \) equal to zero.
• Solving for x, we find \( x = 2 \) and \( x = -\frac{3}{2} \).
• Therefore, the domain of \( g(x) \) is all real numbers except \( x = 2 \) and \( x = -\frac{3}{2} \).

Understanding the domain is crucial before analyzing where the function might be discontinuous.

Discontinuity Points

Discontinuity points are the specific values of x at which a function is not continuous. For rational functions, these occur where the denominator is zero. As seen in the function \( g(x) = \frac{x^{3} + 5x}{(x-2)(2x+3)} \), solving \( (x-2)(2x+3) = 0 \) gives us:

• \( x = 2 \)
• \( x = -\frac{3}{2} \)

These points make the function undefined, causing discontinuities.

To confirm:

• Substitute \( x = 2 \) into the denominator: \( (2-2)(2*2+3) = 0 \)
• Substitute \( x = -\frac{3}{2} \): \( (-\frac{3}{2}-2)(2(-\frac{3}{2})+3) = 0 \)

Thus, \( x = 2 \) and \( x = -\frac{3}{2} \) are confirmed as the discontinuity points.

Vertical Asymptotes

Vertical asymptotes are lines \( x = a \) where a function approaches but never touches or crosses. They occur when the denominator of a rational function is zero, assuming the numerator is non-zero at those points. For the function \( g(x) = \frac{x^{3} + 5x}{(x-2)(2x+3)} \):

• We identified \( x = 2 \) and \( x = -\frac{3}{2} \) as points where the denominator is zero.

These points result in vertical asymptotes. This can be visualized as the graph of \( g(x) \) approaching the lines \( x = 2 \) and \( x = -\frac{3}{2} \) but never touching them. These lines divide the graph into distinct regions reflecting behavior changes.

Rational Functions

Rational functions are of the form \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. The function \( g(x) = \frac{x^{3} + 5x}{(x-2)(2x+3)} \) is a classic example.

Key Characteristics:

• Numerator \( P(x) \): Determines the function values for given inputs.
• Denominator \( Q(x) \): Important for finding domain restrictions. A zero in the denominator means the function is undefined at those points.

When analyzing rational functions:

• Identify where the denominator is zero to find discontinuities and vertical asymptotes.
• Simplify if possible to better understand the function's behavior. For \( g(x) \), discontinuities at \( x = 2 \) and \( x = -\frac{3}{2} \) lead to vertical asymptotes.

Grasping these concepts ensures a deeper understanding of such functions and their behavior.