Problem 75 - Calculus: Early Transcendentals

Question

Sketch the curves. Identify clearly any interesting features, including local maximum and minimum points, inflection points, asymptotes, and intercepts. $$ y=x\left(x^{2}+1\right) $$

Step-by-Step Solution

Answer

Function has no local extrema, an inflection point at \( x = 0 \), and intercept at \( (0,0) \).

Step 1: Simplify the Function

The given function is \( y = x(x^2 + 1) \). Simplify it to get \( y = x^3 + x \). You will work with this equation throughout the following steps.

Step 2: Find the First Derivative

To find critical points, first compute the derivative \( y' \). Differentiating \( y = x^3 + x \) gives \( y' = 3x^2 + 1 \). This derivative helps find local maxima and minima.

Step 3: Find Critical Points

Set the first derivative \( y' = 3x^2 + 1 \) equal to zero to find critical points. Solving \( 3x^2 + 1 = 0 \) gives no real solutions because \( 3x^2 = -1 \) has no real roots. Therefore, there are no local extrema.

Step 4: Find the Second Derivative

To detect inflection points, find the second derivative. Differentiate \( y' = 3x^2 + 1 \) to get \( y'' = 6x \).

Step 5: Solve for Inflection Points

Set the second derivative \( y'' = 6x \) equal to zero to find potential inflection points. Solving \( 6x = 0 \) gives \( x = 0 \). This is a potential inflection point.

Step 6: Verify Inflection Point

Check if the concavity changes at \( x = 0 \). For \( x < 0 \), \( y'' = 6x < 0 \), indicating concavity down. For \( x > 0 \), \( y'' = 6x > 0 \), indicating concavity up. Therefore, \( x = 0 \) is an inflection point.

Step 7: Analyze Asymptotic Behavior

The function \( y = x^3 + x \) is a polynomial and does not have any vertical or horizontal asymptotes because it does not tend to infinity as \( x \to \pm \infty \).

Step 8: Find Intercepts

To find the intercepts, calculate where the function crosses the axes. For the y-intercept, set \( x = 0 \), giving \( y = 0 \). The curve also crosses the x-axis at \( x^3 + x = 0 \), factoring gives \( x(x^2 + 1) = 0 \), so the real root is \( x = 0 \).

Key Concepts

Derivatives

Calculating the derivative of a function is a crucial step in understanding its behavior, particularly when analyzing its slopes and rates of change. For a given function, the first derivative helps identify

• the rate at which the function’s value is changing at any point,
• the existence and location of any local maxima or minima,
• and the points where the slope is zero, indicating a possible change in direction.

For the function given, \( y = x^3 + x \), the first derivative is \( y' = 3x^2 + 1 \). This derivative informs us about the general shape of the graph and how steep it is.
To find critical points, where the function might have peaks or valleys, we set \( y' = 0 \). In this case, \( 3x^2 + 1 = 0 \) leads to no real solutions, indicating there are no local maxima or minima. Therefore, the function’s slope never hits zero nor changes its continuous increase or decrease.
Understanding derivatives is essential in curve sketching to map out these prominent features and confirm their absence or presence within specific sections of a graph.

Inflection Points

Inflection points are spots on the curve where the concavity changes direction. Calculating these points involves using the second derivative, which gives insight into the curvature's nature. For our function,

• find the second derivative, \( y'' = 6x \),
• and then set \( y'' = 0 \) to locate potential inflection points.

In this instance, solving \( 6x = 0 \) yields \( x = 0 \), a possible inflection point. Confirming \( x = 0 \) as an inflection point necessitates checking concavity changes around this point.
For \( x < 0 \), \( y'' \lt \ 0 \), indicating the curve is concave down; for \( x > 0 \), \( y'' \gt \ 0 \), revealing it becomes concave up. This shift in concavity validates \( x = 0 \) as an inflection point, marking a pivotal turn in the curve's behavior.
Recognizing inflection points aids greatly in sketching curves since it pinpoints where the direction of "bending" or "curving" changes on the graph.

Asymptotes

Asymptotes are lines that a graph may approach but never actually touch. They typically illustrate a function’s behavior at significant limits, either as \( x \to \ \ \ \infty \) or \( x \to \ -\infty \). In polynomial functions like \( y = x^3 + x \),

• there are typically no vertical or horizontal asymptotes because the function does not have any restrictions or tendencies to move toward a line infinitely,
• as all terms are polynomial expressions that continue infinitely in both directions.

This function's cubic nature means that as \( x \to \ \ \pm \ \infty \), the values of \( y \) also move towards \( \ \ \pm \ \infty \) without settling into a particular line. Hence, it skips vertical or horizontal asymptotes.
This aspect helps delineate the boundaries and limits of graph behavior without needing to sketch proximity lines that the curve approaches.

Intercepts

Intercepts are crucial points where a graph crosses the x-axis or y-axis. Finding these intercepts is an essential part of curve sketching because they provide fixed points the graph must pass through. For the given function \( y = x^3 + x \), identify

• y-intercepts by setting \( x = 0 \), which results in \( y = 0 \). Thus, the graph passes through the origin.
• x-intercepts by solving \( x^3 + x = 0 \). Factoring the equation gives \( x(x^2 + 1) = 0 \), yielding a real root at \( x = 0 \).

Both solutions confirm that the origin \( (0,0) \) is where the graph intersects both axes.
Understanding where a function intersects the axes enables a clearer graph outline. It serves as a stepping stone for further analytical or graphical investigations.