Problem 29 - Precalculus: Functions and Graphs

Question

Sketch the graph of \(f\) $$f(x)=\frac{x-1}{x^{3}-4 x}$$

Step-by-Step Solution

Answer

The graph of \( f(x) = \frac{x-1}{x^3-4x} \) has vertical asymptotes at \( x = -2, 0, 2 \), a horizontal asymptote at \( y = 0 \), and behaves according to sign changes in defined intervals.

Step 1: Factor the Denominator

First, consider the denominator of the function, \( x^3 - 4x \). Factor out \( x \) to obtain \( x(x^2 - 4) \). The expression \( x^2 - 4 \) can further be factored as \( (x-2)(x+2) \). Thus, the denominator simplifies to \( x(x-2)(x+2) \).

Step 2: Identify Domain Restrictions

The function \( f(x) = \frac{x-1}{x^3 - 4x} \) is undefined where the denominator is zero. Solve for \( x \) in \( x(x-2)(x+2) = 0 \), yielding points \( x = 0, 2, -2 \). Thus, these are vertical asymptotes or holes, depending on the cancellation with the numerator.

Step 3: Simplify the Rational Function

Examine the numerator \( x - 1 \). Since there are no factors of \( x - 1 \) in the denominator, no cancellations occur. The simplified function remains \( f(x) = \frac{x-1}{x(x-2)(x+2)} \).

Step 4: Determine Vertical Asymptotes and Holes

Each factor in the denominator \( x = 0, x = -2, x = 2 \) leads to vertical asymptotes because there is no cancellation with the numerator. Consequently, the vertical asymptotes are at \( x = 0 \), \( x = 2 \), and \( x = -2 \).

Step 5: Identify Horizontal Asymptotes

Evaluate the degrees of the numerator and the denominator. The degree of the numerator is 1, while the degree of the denominator is 3. Since the degree of the denominator is greater than that of the numerator, the horizontal asymptote is \( y = 0 \).

Step 6: Discover the Behavior at Intervals

Examine the sign of \( f(x) \) within intervals determined by the vertical asymptotes \( x = -2, 0, 2 \). Choose test points such as \( x = -3 \), \( x = -1 \), \( x = 1 \), and \( x = 3 \) to determine the sign of \( f(x) \) in these intervals. For example, at \( x = -1 \), the sign across the factors \( (x-1) \), \( x \), \( (x-2) \), and \( (x+2) \) reveals that \( f(-1) = \frac{-1+1}{-1(-1-2)(-1+2)} = - \), indicating \( f(x) < 0 \) in this interval. Repeat this for other test points.

Step 7: Sketch the Graph

Using the established vertical asymptotes \( x = -2, 0, 2 \), horizontal asymptote \( y = 0 \), and the sign analysis for intervals, sketch the function showing the behavior as it approaches asymptotes. Ensure the graph reflects changes in polarity obtained from interval testing.

Key Concepts

Vertical Asymptotes

Vertical asymptotes are vertical lines where a rational function like \( f(x) = \frac{x-1}{x^3 - 4x} \) becomes undefined. This happens when the denominator of the function equals zero, as division by zero is undefined in mathematics.

To find vertical asymptotes, we set the denominator equal to zero and solve for \( x \). In our case, the denominator is \( x(x-2)(x+2) \). Setting each factor equal to zero, we find \( x = 0 \), \( x = 2 \), and \( x = -2 \) as the roots.

Since none of these factors cancel with any term in the numerator, each of these roots indicates a vertical asymptote. Thus, the graph of the function will have vertical asymptotes at these points. The function will tend towards positive or negative infinity as \( x \) approaches these values from either direction.

Here's a simple checklist to find and confirm vertical asymptotes in rational functions:

• Factor the denominator completely.
• Solve the equation by setting it equal to zero to find critical \( x \) values.
• Check for cancellations with the numerator (this would indicate removable discontinuities, not asymptotes).

Horizontal Asymptotes

Unlike vertical asymptotes, horizontal asymptotes describe the behavior of the function \( f(x) = \frac{x-1}{x^3 - 4x} \) as \( x \) approaches infinity or negative infinity. For horizontal asymptotes, we compare the degrees of the numerator and the denominator.

In \( f(x) \), the numerator \( x-1 \) is of degree 1, while the denominator \( x^3 - 4x \) is of degree 3. The rule of thumb is:

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).
• If the degrees are equal, the horizontal asymptote is \( y = \frac{a}{b} \), where \( a \) and \( b \) are the leading coefficients of the numerator and denominator respectively.
• If the degree of the numerator is greater than the denominator, there is no horizontal asymptote.

For our function, since the degree of the denominator is greater than the numerator, the horizontal asymptote is \( y = 0 \). This means that as \( x \) goes to very large positive or negative values, \( f(x) \) will approach the x-axis without ever actually touching it.

Domain Restrictions

Domain restrictions in a rational function identify the values of \( x \) where the function is not defined. For rational functions, this often occurs where the denominator is zero, as division by zero cannot be performed.

To find domain restrictions for \( f(x) = \frac{x-1}{x^3 - 4x} \), set the denominator \( x(x-2)(x+2) \) equal to zero. Solving the equation gives us \( x = 0, 2, \) and \( -2 \). Thus, these are the values where the function will be undefined.

The domain of the function is all real numbers except these critical values. In interval notation, the domain can be expressed as:

• \((-\infty, -2) \cup (-2, 0) \cup (0, 2) \cup (2, \infty)\)

Understanding domain restrictions helps prevent errors when evaluating the function and is crucial for graphing. It's essential to mark these restrictions explicitly when setting up computations and interpreting the behavior around these points.