Problem 39 - Calculus Single Variable

Question

Sketch the graph of each of the functions in Exercise \(25-40,\) exhibiting and labeling: a) all local and globa extrema; b) inflection points; c) intervals on which the func tion is increasing or decreasing; d) intervals on which the function is concave up or concave down; e) all horizontal an vertical asymptotes. $$ f(x)=(13 x+14) /\left(x^{2}-1\right) $$

Step-by-Step Solution

Answer

The function has vertical asymptotes at \(x = -1\) and \(x = 1\), a horizontal asymptote at \(y = 0\), critical and inflection points determined from first and second derivatives, intervals of concavity, convexity, increase, and decrease identified. Sketch reflects these features.

Step 1: Identify the Domain

The function is given by \( f(x) = \frac{13x + 14}{x^2 - 1} \). The denominator \(x^2 - 1\) cannot be zero, so we find where the function is undefined by solving \(x^2 - 1 = 0\). The solutions are \(x = 1\) and \(x = -1\). Therefore, the domain is all real numbers except \(x = 1\) and \(x = -1\).

Step 2: Find the First Derivative

To find the intervals of increase or decrease, calculate the first derivative using the quotient rule: \(f'(x) = \frac{(x^2 - 1)\cdot 13 - (13x + 14)\cdot 2x}{(x^2 - 1)^2}\). Simplify the derivative to find the critical points.

Step 3: Determine Critical Points

Set the first derivative \(f'(x)\) equal to zero and solve for \(x\). This will give you the potential critical points where local maxima or minima may occur.

Step 4: Analyze Interval of Increase/Decrease

Use test points from the intervals determined by the critical points and the domain restrictions to determine where the function is increasing or decreasing. This is decided by the sign of \(f'(x)\). Positive means increasing and negative means decreasing.

Step 5: Find the Second Derivative

To find the concavity and inflection points, calculate the second derivative \(f''(x)\). Use the first derivative you found and apply the derivative rules again.

Step 6: Determine Inflection Points

Set \(f''(x) = 0\) and solve for \(x\) to find potential inflection points. Analyze the sign change of \(f''(x)\) across these points to confirm true inflection points.

Step 7: Analyze Concavity

Using \(f''(x)\), determine the intervals where the function is concave upward (\(f''(x) > 0\)) or concave downward (\(f''(x) < 0\)).

Step 8: Find Asymptotes

Identify vertical asymptotes at the points where the function is undefined, \(x = -1\) and \(x = 1\). To find the horizontal asymptote, evaluate the limits \(\lim_{x \to \pm \infty} f(x)\). The degree of the polynomial in the numerator is less than the denominator, suggesting a horizontal asymptote at \(y = 0\).

Step 9: Sketch the Graph

Using all gathered information, sketch the graph: plot x-values for critical points, inflection points; draw arrows to indicate intervals of increase, decrease, concavity, convexity, and asymptotes. Label all important points and asymptotic lines.

Key Concepts

Local and Global Extrema

To understand local and global extrema, it's important to know that these are points where a function reaches maximum or minimum values within a certain interval or overall. These are critical in graph sketching. Once the function's derivative is set to zero, the solutions are potential minima or maxima. Testing these points can confirm if they are indeed local maxima or local minima.
Local extrema occur where the slope of the tangent line is zero, indicating a peak or trough. However, global extrema are the absolute highest or lowest values of the function over its entire domain.
For example, in the function provided, by solving for critical points in the modified derivative, students can find where these extrema occur. Using intervals and test points helps to differentiate between local and global extrema.

Inflection Points

Inflection points are fascinating as they indicate where a function changes its curvature, switching from convex to concave or vice versa. By finding the second derivative and setting it to zero, students can determine these points. Simply put, these points occur when the rate of change of the slope switches direction.
Once located, these points indeed alter the function's "bend." Hence, analyzing the sign change in the second derivative around these points is crucial for clarity.
In our function, once you compute and examine the second derivative, any resulting zeros provide potential inflection points. Confirm by ensuring the second derivative changes sign around these values.

Increasing and Decreasing Intervals

An essential aspect of graph sketching is understanding where the function increases or decreases. This behavior can be determined using the first derivative.
When the first derivative,

• is positive, the function increases
• is negative, the function decreases

Locating these intervals often involves some calculation and testing.
For the given function, after finding the critical points, it's helpful to test intervals around these points and within the function's domain. Applying test points to the first derivative helps efficiently determine whether the interval increases or decreases.

Concavity and Asymptotes

Understanding concavity enables one to predict how a graph bends. It's determined by the second derivative:

• If the second derivative is positive, the function is concave up.
• If negative, the function is concave down.

This concept is directly related to inflection points, where concavity switches direction.
In addition to concavity, asymptotes play a significant role. Vertical asymptotes occur where the function is undefined, often pointing to infinity. Horizontal asymptotes indicate where the function levels out at extreme values of x.
In the example function, vertical asymptotes are at the undefined values, x = -1 and x = 1, and the function's degree comparison suggests a horizontal asymptote at y = 0. Together, these details carve out the full picture of the function's behavior across its domain.