Problem 31 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 31 through 38 , determine where the given function is increasing and decreasing and where its graph is concave upward and concave downward. Sketch the graph, showing as many key features as possible (high and low points, points of inflection, asymptotes, intercepts, cusps, vertical tangents). 31\. \(f(x)=e^{x}-e^{-x}\)

Step-by-Step Solution

Answer

The function is always increasing. It is concave downward for x < 0 and concave upward for x > 0 with a point of inflection at x = 0.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, find the first derivative of the function. For the function given, : f(x)=e^{x} - e^{-x}, f'(x)= d/dx . e^{x}-e^{-x} = e^{x} + e^{-x}

Step 2: Determine Intervals of Increase and Decrease

Set the first derivative equal to zero to find critical points: e^{x} + e^{-x} = 0 It is apparent that there are no real values for x that satisfy this equation. However, since e^{x} > 0 for all x, e^{x} + e^{-x} is always positive, indicating that f(x) is always increasing.

Step 3: Find the Second Derivative

To determine where the function is concave upward or concave downward, find the second derivative. f''(x)= d/dx (e^{x}+e^{-x}) = e^{x}-e^{-x}

Step 4: Determine Intervals of Concavity

Set the second derivative equal to zero to find points of inflection: e^{x}-e^{-x} = 0 This happens when e^{2x} = 1, . which gives us x = 0. For concavity, if we test points around x = 0, we find: - For x< 0, f''(x) < 0, which means the function is concave downward. - For x > 0, f''(x) > 0, which means the function is concave upward.

Step 5: Identify Key Features

Based on the above steps: - The function is always increasing. - Point of inflection at x = 0. - Concave downward for x < 0 - Concave upward for x > 0 Plotting the function f(x)=e^{x} - e^{-x}, incorporate these details.

Key Concepts

First Derivative Test

The first derivative test helps us determine where a function is increasing or decreasing. We start by finding the first derivative of the given function. For the function \( f(x) = e^{x} - e^{-x} \), the first derivative is:

\( f'(x) = e^{x} + e^{-x} \).

We set the first derivative equal to zero to find the critical points: \( e^{x} + e^{-x} = 0 \). However, there are no real values for \( x \) that satisfy this equation because \( e^{x} > 0 \) for all \( x \). This means \( e^{x} + e^{-x} \) is always positive, indicating that the function is always increasing.

Second Derivative Test

The second derivative test helps analyze the concavity of the graph and identify points of inflection. To start, we find the second derivative:

\( f''(x) = e^{x} - e^{-x} \).

We need to set \( f''(x) \) equal to zero to find points of inflection: \( e^{x} - e^{-x} = 0 \). Solving this, we get \( e^{2x} = 1 \), and thus, \( x = 0 \). This critical point helps us decide intervals of concavity.

Concavity

To determine the intervals where the function is concave upward or downward, we test the sign of the second derivative in the intervals around the critical point \( x = 0 \).

- For \( x < 0 \), \( f''(x) < 0 \), so the function is concave downward.
- For \( x > 0 \), \( f''(x) > 0 \), so the function is concave upward.

This information about concavity tells us how the curve bends in different regions.

Critical Points

Critical points are where the derivative is zero or undefined. They can indicate local maxima, minima, or saddle points. In our function \( f(x) = e^{x} - e^{-x} \), we didn't find any real values where \( f'(x) = 0 \).

This is because the first derivative \( f'(x) = e^{x} + e^{-x} \) is always positive. Hence, the function is always increasing and has no local maxima or minima.

Point of Inflection

A point of inflection is where the concavity of the function changes. This occurs when the second derivative changes sign. For our function, the second derivative \( f''(x) = e^{x} - e^{-x} \) is zero at \( x = 0 \).

- To the left of \( x = 0 \), the function is concave downward.
- To the right of \( x = 0 \), the function is concave upward.

Therefore, \( x = 0 \) is a point of inflection. This key feature is essential when sketching the graph of the function.