Problem 25 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=e^{x}+e^{-x} $$

Step-by-Step Solution

Answer

The function is concave up everywhere, and the local minimum is at \(x = 0\).

Step 1: Find the First Derivative

First, we need to find the first derivative of the function \(f(x) = e^x + e^{-x}\). The derivative of \(e^x\) is \(e^x\), and the derivative of \(e^{-x}\) is \(-e^{-x}\). So, the first derivative, \(f'(x)\), is: \[f'(x) = e^x - e^{-x}\].

Step 2: Solve for Critical Points

Critical points occur where the first derivative equals zero. Solve \(f'(x) = 0\): \[e^x - e^{-x} = 0\] \[e^x = e^{-x}\]. Divide both sides by \(e^x\), resulting in \[1 = e^{-2x}\]. Taking the natural logarithm of both sides gives \[0 = -2x\], solving for \(x\) gives \(x = 0\). So, the critical point is \(x = 0\).

Step 3: Find the Second Derivative

Next, we need to find the second derivative for concavity and the Second Derivative Test. Differentiate \(f'(x) = e^x - e^{-x}\) again to get \(f''(x)\): \[f''(x) = e^x + e^{-x}\].

Step 4: Determine Intervals of Concavity and Points of Inflection

Examine \(f''(x) = e^x + e^{-x}\). Since \(e^x\) and \(e^{-x}\) are always positive for all values of \(x\), \(f''(x)\) is always positive. Therefore, the function is concave up on the entire real line and has no points of inflection.

Step 5: Apply the Second Derivative Test at Critical Points

The Second Derivative Test states that if \(f''(x) > 0\) at a critical point, \(f(x)\) has a local minimum there. Since \(f''(0) = e^0 + e^{-0} = 1 + 1 = 2 > 0\), \(f(x)\) has a local minimum at \(x = 0\).

Key Concepts

First Derivative

To understand concavity and find critical points, the first step is finding the first derivative of a function. For the given function, \(f(x) = e^x + e^{-x}\), the first derivative \(f'(x)\) represents the slopes of tangent lines to the curve of \(f(x)\). Calculating \(f'(x)\) involves differentiating each term separately. The derivative of \(e^x\) remains \(e^x\), and the derivative of \(e^{-x}\) becomes \(-e^{-x}\). Thus, the first derivative is:

• \(f'(x) = e^x - e^{-x}\).

This derivative is crucial for identifying where the function's graph has horizontal tangents, which are typically indicative of potential maxima or minima. These points are named critical points.

Critical Points

Critical points occur at x-values where the first derivative equals zero or is undefined. For the function \(f(x) = e^x + e^{-x}\), critical points arise when \(f'(x) = 0\). Solving \(f'(x) = e^x - e^{-x} = 0\) means:

• \(e^x = e^{-x}\).

Dividing both sides by \(e^x\) results in \(1 = e^{-2x}\). By taking the natural logarithm, we obtain:

• \(0 = -2x\).

Solving for \(x\) gives \(x = 0\), marking it as a critical point. Critical points are significant because they indicate potential locations for local minima or maxima in the function.

Second Derivative Test

Once the critical points are found, the second derivative test helps determine whether these points are local minima, maxima, or neither. The second derivative \(f''(x)\) provides insights into the concavity of the function. For the function \(f(x) = e^x + e^{-x}\), calculating the second derivative results in:

• \(f''(x) = e^x + e^{-x}\).

Both \(e^x\) and \(e^{-x}\) are positive for all real \(x\), making \(f''(x)\) positive everywhere. The Second Derivative Test posits that if \(f''(x) > 0\) at a critical point, the point is a local minimum. At \(x = 0\), \(f''(0) = 2 > 0\) confirms a local minimum at that point.

Local Minimum

A local minimum at a point suggests that in its close vicinity, the function value at that point is lower than those surrounding it. In this exercise, using the function \(f(x) = e^x + e^{-x}\), and focusing on the critical point \(x = 0\), the second derivative test confirmed a local minimum. Since \(f''(0) = 2\), and this value is positive, it proves the point is a local minimum.This means at \(x = 0\), the function has used its given resources efficiently to reach the lowest value in its immediate neighborhood. Understanding local minima is essential for optimization problems and helps in analyzing different real-world scenarios concerning trends, cost efficiencies, and other crucial decisions based on calculus.