Problem 31 - Calculus Early Transcendentals

Question

a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing utility to determine whether each critical point corresponds to a local maximum, local minimum, or neither. $$f(x)=\left(e^{x}+e^{-x}\right) / 2$$

Step-by-Step Solution

Answer

Question: Find the critical points of the function \(f(x) = \frac{e^x + e^{-x}}{2}\) and determine their nature (local maximum, local minimum, or neither) using a graphing utility. Answer: The function has a critical point at \(x=0\) which corresponds to a local minimum. The coordinates of the local minimum point are \((0,1)\).

Step 1: Find the derivative of the function

To find the derivative of the function \(f(x)=\frac{e^{x}+e^{-x}}{2}\), we'll apply the power rule and the chain rule as needed: $$f'(x) = \frac{d}{dx}\left(\frac{e^{x}+e^{-x}}{2}\right)$$ $$= \frac{1}{2}\left(\frac{d}{dx}\left(e^{x}+e^{-x}\right)\right)$$ $$= \frac{1}{2}\left(e^x - e^{-x}\right)$$

Step 2: Find the critical points

To find the critical points, we'll set the derivative, \(f'(x)\), equal to zero and solve for \(x\): $$\frac{1}{2}\left(e^x - e^{-x}\right) = 0$$ We can multiply both sides by 2 to get rid of the fraction: $$(e^x - e^{-x}) = 0$$ Now, let's add \(e^{-x}\) to both sides: $$e^x = e^{-x}$$ To solve for x, we will take the natural logarithm (ln) of both sides: $$x = -x$$ Therefore, the critical point is \(x=0\).

Step 3: Determine nature of the critical points

Our critical point is \(x = 0\). In this step, we'll use a graphing utility to evaluate the function at the critical point and determine the nature (local maximum, local minimum, or neither) of this point. $$f(x) = \frac{e^x + e^{-x}}{2}$$ $$f(0) = \frac{e^0 + e^{-0}}{2} = \frac{1+1}{2} = 1$$ When using a graphing utility (such as Desmos, GeoGebra, or a graphing calculator), we can visualize that when \(x=0\), the function has a local minimum value at the point \((0,1)\) as the graph is concave up at this point.

Step 4: Final answer

The function \(f(x)=\frac{e^{x}+e^{-x}}{2}\) has a critical point at \(x=0\) which corresponds to a local minimum. The coordinates of the local minimum point are \((0,1)\).

Key Concepts

Derivative of Exponential Functions

Understanding the derivative of exponential functions is crucial in calculus as these functions appear frequently in various applications. Exponential functions are functions of the form f(x) = a^x, where a is a constant and x is the variable. However, the most common base for an exponential function in calculus is the natural base e.

The derivative of an exponential function with base e is unique because the rate of change of the function at any point is equal to the value of the function at that point. This is due to the constant e's unique properties. Using the rules of differentiation, specifically the chain rule, the derivative of e^x is simply e^x, and the derivative of e^{-x} is -e^{-x}.

When we look at a function like f(x) = (e^x + e^{-x}) / 2, finding the derivative involves applying the sum and constant multiple rules as well. The function's derivative f'(x) gives us the slope of the tangent line at any point x on the curve, which is essential for identifying the behavior of the function, including the location and nature of critical points.

Local Minimum and Maximum

Local minimum and maximum points are often referred to as local extrema. They represent points on a graph where the function reaches a peak or valley in a particular, localized area. These points are essential in understanding the function's overall behavior and optimizing real-world problems such as cost or efficiency.

To find local extrema, we first calculate the derivative of the function to find where the slope of the tangent is zero, that is, where f'(x) = 0. These points are potential candidates for local minima or maxima. However, finding the derivative equal to zero provides us with critical points, but does not guarantee that all critical points are local extrema.

We use different tests, like the first or second derivative tests, to determine the nature of these critical points. In the case of f(x) = (e^x + e^{-x}) / 2, after finding the critical point at x = 0, we can analyze the concavity, or the second derivative, at that point to determine if it is a local minimum or maximum. If the function curves upward, as it does in this example, it indicates a local minimum.

Graphing Utilities in Calculus

Graphing utilities, which include graphing calculators and software like Desmos or GeoGebra, are invaluable tools for visualizing functions and their properties. They provide a dynamic way to analyze the graph of a function, enabling students to immediately see the effects of changes in the equation on the graph.

When we use these utilities to investigate functions, we can visually identify behaviors like increasing and decreasing intervals, symmetry, and pinpoint local extrema with great accuracy. For more intricate functions where algebraic methods may be challenging, graphing utilities can provide a clear understanding of the function's shape and behavior.

For instance, when analyzing the function f(x) = (e^x + e^{-x}) / 2, we can graph it and observe the concavity around the critical point. If the graph appears as a 'U' shape around the critical point, it indicates a local minimum. Graphing utilities are especially helpful when approaching complex functions, as they assist in confirming the results determined analytically and providing a visual representation to deepen our understanding of calculus concepts.