Problem 63 - University Calculus: Early Transcendentals

Question

Find the extreme values (absolute and local) of the function over its natural domain, and where they occur. $$y=e^{x}+e^{-x}$$

Step-by-Step Solution

Answer

The function has a local minimum at \( x = 0 \) with value \( 2 \). There is no absolute maximum or minimum.

Step 1: Differentiate the Function

To find the extreme values, we first need to find the derivative of the function. The given function is \( y = e^x + e^{-x} \). Differentiate it with respect to \( x \):\[y' = \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{-x}) = e^x - e^{-x}.\]

Step 2: Find Critical Points

Set the derivative equal to zero to find the critical points:\[e^x - e^{-x} = 0.\]This can be rewritten as:\[e^x = e^{-x}.\]Multiply both sides by \( e^x \):\[(e^x)^2 = 1 \Rightarrow e^{2x} = 1.\]Solving for \( x \), we get:\[2x = 0 \Rightarrow x = 0.\]Thus, the critical point is \( x = 0 \).

Step 3: Determine the Nature of Critical Points

To determine whether the critical point is a local maximum, minimum, or neither, use the second derivative test. First, find the second derivative:\[y'' = \frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x}.\]Evaluate the second derivative at \( x = 0 \):\[y''(0) = e^0 + e^0 = 1 + 1 = 2.\]Since \( y''(0) > 0 \), the critical point at \( x = 0 \) is a local minimum.

Step 4: Evaluate the Function at Critical Points and Endpoints

Since the function \( y = e^{x} + e^{-x} \) is defined for all real numbers and there are no finite endpoints, we only need to evaluate the function at the critical point. Calculate the value of the function at \( x = 0 \):\[y(0) = e^0 + e^0 = 1 + 1 = 2.\]So, \( y = 2 \) is the value at the local minimum.

Step 5: Analyze Behavior at Infinity

Examine the function as \( x \to \pm\infty \). As \( x \to \infty \), \( e^x \) grows without bound, and \( e^{-x} \) approaches 0, so \( y \to \infty \). As \( x \to -\infty \), \( e^x \to 0 \) and \( e^{-x} \to \infty \), so \( y \to \infty \). Thus, the function has no absolute maximum or minimum because it increases without bounds in both directions.

Key Concepts

Extreme Values

When dealing with functions, extreme values are the highest or lowest points in a given range. These can be classified as absolute (global) or local extremes. An absolute extreme is the highest or lowest point across the entire domain of the function, while a local extreme is the highest or lowest point within a neighborhood surrounding a specific point.

In our example, the function is defined as \( y = e^x + e^{-x} \) and is continuous over all real numbers. To find the extreme values of this function, we look for the places where the slope changes, that is, where the derivative is zero or undefined (critical points) and evaluate the function's behavior around them.

No absolute extreme values exist here, as the function grows infinitely in both positive and negative directions. We do find a local minimum at \( x = 0 \), as the function reaches its lowest point in that specific region.

Derivative

To identify the critical points of a function, one must first calculate its derivative. The derivative represents the rate at which the function is changing at any point, which is essential for determining the extremity of values.

For our function \( y = e^x + e^{-x} \), the derivative is found using the rules of differentiation:

• The derivative of \( e^x \) is \( e^x \).
• The derivative of \( e^{-x} \) is \( -e^{-x} \).

Therefore, the derivative is \( y' = e^x - e^{-x} \).
With this information, critical points can be found by setting the derivative to zero, signifying points where the function's slope is horizontal.

Critical Points

Critical points occur where the derivative of the function is zero or undefined. These points are potential candidates for extreme values, as they represent locations where the function's rate of change is zero.

In our problem, the derivative \( y' = e^x - e^{-x} \) is set to zero:

• Solving \( e^x - e^{-x} = 0 \) gives us \( e^x = e^{-x} \).
• After rearranging and solving, we find the critical point at \( x = 0 \).

This critical point is an important place to evaluate for finding extreme values, and further analysis with the second derivative will help determine the nature of this extreme—whether it is a maximum, minimum, or neither.

Second Derivative Test

The second derivative test helps determine the nature of the critical points found using the first derivative. This test involves computing the second derivative and evaluating it at the critical points.

For the function \( y = e^x + e^{-x} \), the second derivative is:

• \( y'' = e^x + e^{-x} \).
• By substituting the critical point \( x = 0 \) into the second derivative, we calculate \( y''(0) = 2 \).

Since \( y''(0) > 0 \), this indicates that the function is concave up at \( x = 0 \), confirming a local minimum.

This test provides a straightforward way to conclude the nature of critical points and can be a more intuitive approach compared to other methods for understanding the function's curvature and behavior.