Problem 9 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 through 10 , determine intervals of increase and decrease and intervals of concavity for the given function. Then sketch the graph of the function. Be sure to show all key features such as intercepts, asymptotes, high and low points, points of inflection, cusps, and vertical tangents. $$ F(x)=2 x+\frac{8}{x}+2 $$

Step-by-Step Solution

Answer

Intervals of increase: \( x > 0 \). Intervals of concavity: concave down \( x \to -\infty \); concave up \( x \to \infty \). Vertical asymptote at \( x = 0 \).

Step 1 - Find the First Derivative

Compute the first derivative of the function to determine the intervals of increase and decrease. Given the function \[ F(x)=2x+\frac{8}{x}+2 \], the first derivative using the power rule and the derivative of a reciprocal is \[ F'(x) = 2 - \frac{8}{x^2} \].

Step 2 - Solve for Critical Points

Set the first derivative equal to zero and solve for x to find critical points: \[ 2 - \frac{8}{x^2} = 0 \]. Solving for x, we get \[ 2 = \frac{8}{x^2} \] which simplifies to \[ x^2 = 4 \] and gives the critical points \[ x = 2 \] and \[ x = -2 \].

Step 3 - Determine Intervals of Increase and Decrease

Test intervals around the critical points to determine where the function is increasing or decreasing. Pick test points in each interval \( (-\infty, -2) \), \( (-2, 2) \), and \( (2, \infty) \):1. For \( x = -3 \) in \( (-\infty, -2) \): \( F'(-3) = 2 - \frac{8}{9} > 0 \), increasing. 2. For \( x = 0 \) in \( (-2, 2) \): \( F'(0) \) is undefined, so not applicable. 3. For \( x = 3 \) in \( (2, \infty) \): \( F'(3) = 2 - \frac{8}{9} > 0 \), increasing.

Step 4 - Find the Second Derivative

Compute the second derivative to find intervals of concavity. From \( F'(x) = 2 -\frac{8}{x^2} \), the second derivative is \[ F''(x) = \frac{16}{x^3} \].

Step 5 - Determine Concavity

Analyze the sign of the second derivative to find concavity. Considering \( x^3 \) in the denominator: 1. \( (-\infty, 0) \): \( F''(x) < 0 \), concave down. 2. \( (0, \infty) \): \( F''(x) > 0 \), concave up.

Step 6 - Find Points of Inflection

Identify points where \( F''(x) = 0 \) or changes sign. Since \( F''(x) \) switches signs between \( (-\infty, 0) \) and \( (0, \infty) \), the point of inflection is at \( x = 0 \). However, \( F(x) \) is undefined at \( x = 0 \).

Step 7 - Identify Intercepts

Find the x- and y-intercepts. Set \( F(x) = 0 \) to find x-intercepts and evaluate \( F(0) \) for the y-intercept. Since \( F(x) = 2x + \frac{8}{x} + 2 \), x-intercept calculation is complex and iterative solving might be required. Y-intercept does not exist as \( F(0) \) is undefined.

Step 8 - Evaluate Asymptotes

Analyze vertical and horizontal asymptotes. Vertical asymptote occurs at \( x = 0 \) since \( \frac{8}{x} \) becomes undefined. As \( x \to \infty \) or \( x \to -\infty \), \( F(x) \approx 2x \) (linear behavior).

Step 9 - Sketch the Graph

Combine all previous steps to sketch the graph. Include intercepts, asymptotes, intervals of increase/decrease, and concavity observations.

Key Concepts

First Derivative

To understand the behavior of the function and its intervals of increase and decrease, the first derivative is crucial. For the function To analyze the critical points, solve it by setting Now for It helps understand regions where the function is increasing or decreasing. By plugging in values in intervals divided by the critical points (.

Critical Points

Critical points occur where the first derivative is zero or undefined. For our function, critical points were found by solving the equation These points ( These critical points help in identifying significant changes in the graph, like potential maximum or minimum values. By testing intervals around them, it's possible to determine if the function is increasing or decreasing in those regions.

Second Derivative

The second derivative provides information on the concavity of the function. It is derived from the first derivative and is central in determining points of inflection. For this function, The second derivative was found to be The sign of this second derivative on intervals around concavity.

Concavity

Concavity describes the curvature of the graph. It tells if the graph is curving upward or downward. Using the second derivative, we can determine concavity by examining its sign:

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  Considering various intervals, for x-values in back down for intervals where indicating concavity upwards. Simple and clear visualization aids in sketching accurate diagrams.

Intercepts

Intercepts give key points where the graph intersects the axes. Specifically, x-intercepts occur when find these points.
The y-intercept could be found by substituting coordinate would be on the y-axis. However, the function well as the denominator's presence.

Recognizing intercepts is essential in laying a meaningful graph.

Asymptotes

Asymptotes are lines that the graph approaches but never touches. Vertical asymptotes generally occur where the function becomes undefined, like a division by zero. For therefore making it a vertical asymptote.
Horizontal asymptotes can often be found by examining behavior as functions' comparative nature at extreme values helps predict this.
Including asymptotes in a graph helps showcase behavior limits of the function.