Problem 24 - Calculus Early Transcendentals

Question

Sketch a graph of the rational function and label the coordinates of the stationary points and inflection points. Show the horizontal, vertical, oblique, and curvilinear asymptotes and label them with their equations. Label point(s), if any, where the graph crosses an asymptote. Check your work with a graphing utility. $$ \frac{x^{5}}{x^{2}+1} $$

Step-by-Step Solution

Answer

No vertical or horizontal asymptotes; oblique asymptote at \(y = x^3\); critical point at \((0, 0)\); use graphing utility for verification.

Step 1: Identify the Rational Function

The given rational function is \( f(x) = \frac{x^5}{x^2 + 1} \). This function is already expressed in the form of a rational function, where the numerator is a polynomial of degree 5 and the denominator is a polynomial of degree 2.

Step 2: Determine Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero. However, since \( x^2 + 1 \) is always positive (as \( x^2 \geq 0 \)), this function has no vertical asymptotes.

Step 3: Determine Horizontal Asymptotes

To find the horizontal asymptotes, consider the degree of the polynomial in the numerator \(5\) and the denominator \(2\). Since the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes.

Step 4: Determine Oblique Asymptotes

For a rational function where the degree of the numerator is exactly one more than that of the denominator, you find an oblique asymptote by performing polynomial long division. Dividing \(x^5\) by \(x^2 + 1\) gives the quotient \(x^3\), which is the equation of the oblique asymptote.

Step 5: Calculate Critical Points

Differentiate \(f(x)\) to find \(f'(x)\). Using the quotient rule: \(f'(x) = \frac{(5x^4)(x^2 + 1) - (x^5)(2x)}{(x^2 + 1)^2} = \frac{5x^6 + 5x^4 - 2x^6}{(x^2 + 1)^2} = \frac{3x^6 + 5x^4}{(x^2 + 1)^2}\).Set \(f'(x) = 0\) to find critical points: \(3x^6 + 5x^4 = 0\), which simplifies to \(x^4(3x^2 + 5) = 0\). Solving gives \(x = 0\).

Step 6: Calculate Inflection Points

To find inflection points, differentiate \(f'(x)\) to get \(f''(x)\) and set \(f''(x) = 0\). The second derivative is quite complex, so we focus on where the concavity changes by solving this derivative equation for changes in sign.

Step 7: Identify Points of Intersection

To find points where the graph intersects the asymptotes, set \(f(x) = x^3\) and solve for \(x\). Equating gives \(x^3 = \frac{x^5}{x^2 + 1}\), which simplifies to \(x^4 = x^2 + 1\). Solving this equation provides the intersection points.

Step 8: Graph the Function

Using the information calculated, plot the graph of \(f(x) = \frac{x^5}{x^2 + 1}\). Mark and label the oblique asymptote \(y = x^3\), the critical point at \((0, 0)\), and any intersection points with the asymptotes.

Step 9: Verify with Graphing Utility

Using a tool like Desmos or a graphing calculator, input \(f(x) = \frac{x^5}{x^2 + 1}\) to confirm the sketch. Check that all critical points, asymptotes, and inflection points match the manually calculated values and positions.

Key Concepts

Understanding Asymptotes

Asymptotes are lines that a graph approaches but never actually touches. They are crucial for understanding the behavior of rational functions, particularly at extreme values of the independent variable, often denoted as "x." In the given exercise, we deal with vertical, horizontal, and oblique asymptotes.

• Vertical Asymptotes: These occur where the denominator of a rational function becomes zero, causing the function to approach infinity. For our function \( f(x) = \frac{x^5}{x^2 + 1} \), the denominator \(x^2 + 1\) never equals zero since \(x^2 \geq 0\). Therefore, there are no vertical asymptotes.
• Horizontal Asymptotes: These asymptotes describe the end behavior of a function as \(x\) approaches infinity. Look at the degrees of the numerator and denominator. Here, the numerator's degree (5) is higher, indicating that there are no horizontal asymptotes.
• Oblique Asymptotes: These occur when the degree of the numerator is exactly one more than that of the denominator. By dividing \(x^5\) by \(x^2 + 1\), we obtain the oblique asymptote \(y = x^3\).

Keep these concepts in mind when dealing with rational functions. They help predict the shape and direction of the graph, even without precise drawing.

Uncovering Critical Points

Critical points of a function are where its derivative is zero or undefined. They are important as potential locations for local maxima or minima. Let's explore how to find and interpret these for our given function.

First, we differentiate our function \( f(x) = \frac{x^5}{x^2 + 1} \) using the quotient rule, which gives us \( f'(x) = \frac{3x^6 + 5x^4}{(x^2 + 1)^2} \). By setting the derivative equal to zero, \(3x^6 + 5x^4 = 0\), we solve for the critical points. This simplifies to \(x^4(3x^2 + 5) = 0\).

• Solving the Simplified Equation: We find \(x = 0\), which is our critical point. Evaluating \(f(0)\) gives us \((0, 0)\), a point on the graph.

This critical point can indicate a potential change in the direction of the graph. Always consider these points when analyzing the behavior and shape of rational functions.

Exploring Inflection Points

Inflection points are where a function changes concavity, shifting from curving upward to downward or vice versa. These points are found by examining the second derivative of the function.

For our function \( f(x) = \frac{x^5}{x^2 + 1} \), the inflection points are determined by evaluating the second derivative \( f''(x) \). Solving \( f''(x) = 0 \) can be complex, but it is key to finding where concavity changes.

• Finding Concavity Changes: By examining \( f''(x) \), you can determine intervals where the graph transitions from concave up to concave down.

Although this may seem advanced, understanding where and why the graph curves can significantly enhance sketching activities and improve comprehension of the function's overall shape.