Problem 2 - Calculus Early Transcendentals

Question

Give a graph of the rational function and label the coordinates of the stationary points and inflection points. Show the horizontal and vertical asymptotes and label them with their equations. Label point(s), if any, where the graph crosses a horizontal asymptote. Check your work with a graphing utility. $$ \frac{8}{x^{2}-4} $$

Step-by-Step Solution

Answer

Vertical asymptotes: \( x = \pm 2 \); horizontal asymptote: \( y = 0 \); stationary point: \( (0, -2) \).

Step 1: Identify Discontinuities

The given rational function is \( f(x) = \frac{8}{x^2 - 4} \). The function is undefined where the denominator equals zero. Solve \( x^2 - 4 = 0 \): Factor as \( (x-2)(x+2)=0 \), giving \( x = 2 \) and \( x = -2 \). These values indicate the vertical asymptotes.

Step 2: Find Horizontal Asymptote

For rational functions \( \frac{N(x)}{D(x)} \), if the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \). Here, \( N(x) = 8 \) (degree 0) and \( D(x)=x^2-4 \) (degree 2), so the horizontal asymptote is \( y = 0 \).

Step 3: Calculate Derivative for Stationary Points

Find \( f'(x) \) to locate stationary points. Use the quotient rule: \( f'(x) = \frac{(0 \cdot (x^2 - 4)) - (8 \cdot 2x)}{(x^2 - 4)^2} = \frac{-16x}{(x^2 - 4)^2} \). Set \( f'(x) = 0 \) to find x-values. Since the numerator \( -16x = 0 \) when \( x = 0 \), \( x = 0 \) is the only stationary point. Determine \( f(0) = \frac{8}{0^2-4} = -2 \), so the stationary point is \( (0, -2) \).

Step 4: Calculate Second Derivative for Inflection Points

Use the chain and quotient rule to find \( f''(x) = \frac{d}{dx}\left(\frac{-16x}{(x^2-4)^2}\right) \). Apply differentiation rules: \( f''(x) = \frac{(32(x^2-4)^2 - (-16x)(4x(x^2-4)))}{(x^2-4)^4} \). Simplify and set \( f''(x)=0 \) to solve for inflection points. This typically requires graphing to accurately determine, as the expression is complex.

Step 5: Graph the Function

Plot the function using the information derived. Include vertical asymptotes at \( x = 2 \) and \( x = -2 \), a horizontal asymptote at \( y = 0 \), and label the stationary point \( (0, -2) \). Use a graphing utility, like Desmos or a graphing calculator, to visualize and confirm these points and any additional inflection points.

Step 6: Verify with a Graphing Utility

Input \( f(x) = \frac{8}{x^2 - 4} \) into a graphing utility to evaluate accuracy. Confirm all asymptotes, stationary points, and possible inflection points identified in previous steps.

Key Concepts

Vertical Asymptotes

A vertical asymptote in a rational function is a vertical line where the function tends towards infinity. It occurs at the values of x where the denominator equals zero, causing the function to be undefined.
For the function given, \( f(x) = \frac{8}{x^2 - 4} \), we find the vertical asymptotes by setting the denominator \(x^2 - 4 = 0\).
**Solving for Vertical Asymptotes:**

• Factor the denominator: \((x-2)(x+2)=0\)
• Solve for x: \(x = 2\) and \(x = -2\)

Thus, the vertical asymptotes are at \(x = 2\) and \(x = -2\). As you approach these x-values, the function value increases or decreases without bound. This behavior should be visible on the graph.

Horizontal Asymptotes

Horizontal asymptotes indicate the value that the function approaches as \(x\) approaches infinity or negative infinity.For a rational function \( \frac{N(x)}{D(x)} \), where \(N(x)\) and \(D(x)\) are polynomials:

• 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, it's \( \frac{leading\, coefficient \; of\; N(x)}{leading \; coefficient \; of \; D(x)} \).

In the function \( f(x) = \frac{8}{x^2 - 4} \), the degree of the numerator \(N(x) = 8\) is 0, while the degree of the denominator \(D(x)=x^2-4\) is 2. Therefore, the horizontal asymptote is \(y = 0\).
As \(x\) moves towards positive or negative infinity, the function value approaches zero.

Stationary Points

Stationary points occur where the derivative of the function is zero. These points may represent local maxima, minima, or saddle points.To find the stationary points, calculate the derivative \(f'(x)\) using the quotient rule.The derivative is:\[f'(x) = \frac{-16x}{(x^2 - 4)^2}\]Set \(f'(x) = 0\) to find the x-values where the stationary points occur. Since \(-16x = 0\) solves to \(x=0\), this gives a stationary point.
**Determine the Stationary Point:**

• Compute \(f(0) = \frac{8}{0^2-4} = -2\)
• Thus, the stationary point is \((0, -2)\)

At this point, the slope of the tangent to the curve is zero, indicating a possible turning point.

Inflection Points

Inflection points are found where the second derivative of the function changes sign, indicating a change in concavity.To determine inflection points, calculate the second derivative \( f''(x) \). The second derivative is:\[f''(x) = \frac{(32(x^2-4)^2 - (-16x)(4x(x^2-4)))}{(x^2-4)^4}\]While it's complex to analytically solve \(f''(x)=0\) without graphing, graphing utilities aid in finding where the sign changes. This changes from concave up to concave down or vice versa.**Finding Inflection Points:**

• Check where \(f''(x)\) changes sign.
• These are the approximate x-values of inflection points.

When graphing, look for bends that change curvature. This visualization confirms analytical findings and offers a complete view of the function's behavior.