Problem 10 - 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. $$ f(x)=\frac{1}{x^{3}}+\frac{2}{x^{2}}+\frac{1}{x} $$

Step-by-Step Solution

Answer

The function is always decreasing and has a concavity change at \(x = 0\).

Step 1 - Derive the First Derivative

Find the first derivative of the given function to determine the intervals of increase and decrease. Use the power rule for differentiation:\[ f(x)=\frac{1}{x^{3}}+\frac{2}{x^{2}}+\frac{1}{x} \]The first derivative of the function is: \[ f'(x) = -\frac{3}{x^4} - \frac{4}{x^3} - \frac{1}{x^2} \]

Step 2 - Critical Points and Intervals of Increase/Decrease

Set the first derivative equal to zero and solve for critical points:\[ -\frac{3}{x^4} - \frac{4}{x^3} - \frac{1}{x^2} = 0 \]There are no real solutions for this equation. So, consider the sign of the first derivative to identify intervals of increase and decrease:For \(x > 0\), \(f'(x) < 0\), the function is decreasing. For \(x < 0\), \(f'(x) < 0\), the function is also decreasing.

Step 3 - Derive the Second Derivative

Find the second derivative of the function to determine the intervals of concavity:\[ f''(x) = \frac{12}{x^5} + \frac{12}{x^4} + \frac{2}{x^3} \]

Step 4 - Points of Inflection and Intervals of Concavity

Set the second derivative equal to zero and solve for points of inflection:\[ \frac{12}{x^5} + \frac{12}{x^4} + \frac{2}{x^3} = 0 \]There are no real solutions for this equation. So, consider the sign of the second derivative to identify intervals of concavity:For \(x > 0\), \(f''(x) > 0\), the function is concave up. For \(x < 0\), \(f''(x) < 0\), the function is concave down.

Step 5 - Sketch the Graph

Plot the function using the identified characteristics. Note the intercepts, asymptotes, high and low points, and points of inflection:- The function tends to infinity as \(x\) approaches 0 from either side (vertical asymptote at \(x = 0\)).- The function is decreasing for all \(x\) except at \(x = 0\).- The function is concave up for \(x > 0\) and concave down for \(x < 0\).- Key points: Intercept at \(x = 1\) with value \(f(1) = 4\).

Key Concepts

Intervals of Increase and Decrease

Understanding intervals of increase and decrease helps us analyze a function's behavior. To find these intervals, we use the first derivative, which tells us where the function is increasing or decreasing. Let's look at the given function: \[ f(x) = \frac{1}{x^3} + \frac{2}{x^2} + \frac{1}{x} \]The first derivative is: \[ f'(x) = -\frac{3}{x^4} - \frac{4}{x^3} - \frac{1}{x^2} \] To find where the function increases or decreases, we set the first derivative equal to zero and solve for critical points. However, in this example, there are no real solutions. Therefore, we need to look at the sign of the derivative in different intervals. We find:

• For \( x > 0 \), \( f'(x) < 0 \), so the function is decreasing.
• For \( x < 0 \), \( f'(x) < 0 \), so the function is also decreasing.

Both sides of the graph are decreasing, showing no intervals of increase in this case.

Intervals of Concavity

Intervals of concavity reveal where the function curves upward or downward. We use the second derivative for this analysis. The function's second derivative is:\[ f''(x) = \frac{12}{x^5} + \frac{12}{x^4} + \frac{2}{x^3} \] Just as with the first derivative, we set the second derivative to zero to find inflection points. Here, we find no real solutions either. So then, we analyze the sign of the second derivative:

• For \( x > 0 \), \( f''(x) > 0 \), the function is concave up.
• For \( x < 0 \), \( f''(x) < 0 \), the function is concave down.

Thus, the function changes concavity around the vertical asymptote at \( x = 0 \), curving upwards for positive \( x \) and downwards for negative \( x \).

Calculus Derivatives

Calculus derivatives are the backbone of analyzing functions. The first derivative \( f'(x) \) gives us the slope of the function, indicating where it increases or decreases. For example, \[ f'(x) = -\frac{3}{x^4} - \frac{4}{x^3} - \frac{1}{x^2} \] shows us that the slope is always negative, hence the function is always decreasing.The second derivative \( f''(x) \) shows the rate at which the slope changes. This reveals how the function curves, known as concavity. Here, \[ f''(x) = \frac{12}{x^5} + \frac{12}{x^4} + \frac{2}{x^3} \] indicates concavity. For \( x > 0 \), the positive sign means the curve is upwards (concave up), and for \( x < 0 \), the negative sign means the curve is downwards (concave down).Understanding both derivatives aids in sketching the graph accurately. This helps us capture all key features like intercepts, asymptotes, high and low points, and inflection points.