Problem 14 - 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. $$ \frac{x^{2}}{1-x^{3}} $$

Step-by-Step Solution

Answer

Plot with vertical asymptote at \(x=1\), horizontal at \(y=0\). Stationary points at \((0,0)\) and \((-\sqrt[3]{2},f(-\sqrt[3]{2}))\).

Step 1: Determine the Domain and Identify Asymptotes

The function is \( f(x) = \frac{x^2}{1-x^3} \). Start by identifying the vertical asymptotes by setting the denominator equal to zero: \(1-x^3=0\). Solving this gives \(x^3=1\), so \(x=1\). The rational function \(\frac{x^2}{1-x^3}\) has a vertical asymptote at \(x=1\). Next, identify the horizontal asymptote by considering the degrees of the numerator and the denominator. The degree of the numerator is 2, and the degree of the denominator is 3. Since the degree of the denominator is greater, the horizontal asymptote is at \(y=0\).

Step 2: Find Stationary Points

Stationary points occur where the derivative is zero. Find the derivative of \(f\) using the quotient rule: \[ f'(x) = \frac{(2x)(1-x^3) - (x^2)(-3x^2)}{(1-x^3)^2} \].Simplify the derivative:\[ f'(x) = \frac{2x - 2x^4 + 3x^4}{(1-x^3)^2} = \frac{2x + x^4}{(1-x^3)^2} \].Set \( f'(x) = 0 \) to find critical points:\[ 2x + x^4 = 0 \].Factor out \(x\):\[ x(2 + x^3) = 0 \].Thus, \(x=0\) or \(x = -\sqrt[3]{2}\). Find \(y\)-coordinates by substituting back into \(f(x)\). At \(x=0\), \(f(0)=0\). At \(x=-\sqrt[3]{2}\), substitute to find \(y\).

Step 3: Find Inflection Points

Inflection points occur where the second derivative changes sign. Compute the second derivative by differentiating the first derivative: Using the quotient rule on: \[ f'(x) = \frac{2x + x^4}{(1-x^3)^2} \].The calculation is complex and requires setting the second derivative equal to zero and solving for \(x\). After finding possible values, test intervals to check sign changes. Determine \(y\)-values by substituting into \(f(x)\).

Step 4: Plot the Function and Label Features

Plot the function \(f(x) = \frac{x^2}{1-x^3}\). Label the vertical asymptote at \(x=1\) and the horizontal asymptote at \(y=0\). Mark and label stationary points, inflection points, and where the curve may cross the horizontal asymptote by considering its \(x\)-intercepts. Verify with a graphing utility to ensure accurate representation.

Key Concepts

Asymptotes

In rational functions, asymptotes are lines that the graph of the function approaches but never actually touches or crosses in most cases. There are two primary types of asymptotes: vertical and horizontal.

For the function \[ f(x) = \frac{x^2}{1-x^3} \]vertical asymptotes occur where the denominator equals zero. This means solving\[1-x^3=0\],giving us a vertical asymptote at \(x=1\).

Horizontal asymptotes are determined by comparing the degrees of the polynomials in the numerator and denominator. When the degree of the denominator is higher, as it is here (degree 3 in the denominator compared to degree 2 in the numerator), the horizontal asymptote is\( y=0 \).These lines help shape the graph and provide insight into its behavior as x approaches positive or negative infinity.

Remember, the graph can cross a horizontal asymptote, and identifying these crossings can assist in understanding the overall graph setup.

Stationary Points

Stationary points in a function occur where the derivative equals zero. At these points, the slope of the tangent is flat, meaning the function is neither increasing nor decreasing.

For the function \( f(x) = \frac{x^2}{1-x^3} \),use the quotient rule to find the derivative:\[ f'(x) = \frac{(2x)(1-x^3) - (x^2)(-3x^2)}{(1-x^3)^2} = \frac{2x + x^4}{(1-x^3)^2} \]By setting \( f'(x) = 0 \), you identify critical points \( x=0 \)and \( x=-\sqrt[3]{2} \).

At \( x=0 \), the function value is \( f(0)=0 \).For \(x=-\sqrt[3]{2}\),inserting back into \(f(x)\) gives its corresponding y-value.These points are critical in determining where the graph changes its direction or establishes a flat tangent.

Inflection Points

Inflection points are locations on the graph where the curve changes concavity. This change happens where the second derivative shifts its sign (from positive to negative, or vice versa).

To find these points for \( f(x) = \frac{x^2}{1-x^3} \),calculate the second derivative by differentiating the first derivative again. Use the quotient rule on \( f'(x) = \frac{2x + x^4}{(1-x^3)^2} \).

Solving \( f''(x) = 0 \)will give potential inflection points. Test these points further by checking the intervals on either side to confirm a sign change. Substitute these critical points back into \( f(x) \)to find the corresponding y-values.These inflection points help show where the graph switches from upward concave ("smiling") to downward concave ("frowning"), providing essential clues to the curve's behavior.