Problem 10 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 1 through 12: (a) Find all critical points. (b) Find \(f^{\prime \prime} .\) Use the second derivative, wherever possible, to determine which critical points are local maxima and which are local minima. If the second derivative test fails or is inapplicable, explain why and use an alternative method for classifying the critical point.$$ f(x)=x e^{x}-e^{x} $$

Step-by-Step Solution

Answer

The function \(f(x) = xe^x - e^x\) has a local minimum at \(x = 0\)

Step 1: Compute the first derivative

The first derivative of a function is found by applying the derivative rules. Here the product rule is used. The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. So the derivative of \(f(x) = xe^x - e^x\) is\(f'(x) = (1)(e^x) + (x)(e^x) - (1)(e^x) = xe^x \).

Step 2: Find the critical points

Critical points are where the derivative of a function equals zero or is undefined. To find these for \(f(x)\), set the first derivative equal to zero and solve for \(x\):\(xe^x = 0 \)separating the terms:\( x = 0 \). So \(0\) is the critical point.

Step 3: Compute the second derivative

The second derivative tests the concavity of a function. If it is positive then the function has a local minimum and if it is negative then it has a local maximum. The second derivative of \(f(x)\) is:\(f''(x) = (1)(e^x) + (1)(e^x) + (x)(e^x) = (2 + x)e^x \).

Step 4: Classify the critical points

To classify the critical points, plug them into the second derivative. Therefore at \(x = 0\), \(f''(0) = (2 + 0)(e^0) = 2.This value is positive, so the critical point at \(x = 0\) is a local minimum.

Key Concepts

Derivative Rules

Derivatives are a fundamental part of calculus, used to measure how a function's output changes as its input changes. Learning the correct rules to differentiate functions is essential. Here are some basic rules:

• Constant Rule: The derivative of a constant is zero.
• Power Rule: For any function of the form \(f(x) = x^n\), the derivative is \(f'(x) = nx^{n-1}\).
• Sum Rule: The derivative of a sum is the sum of the derivatives.
• Difference Rule: The derivative of a difference is the difference of the derivatives.

For more complex functions, like in our original exercise, the Product Rule comes into play. Often, combining multiple derivative rules helps find the derivative of a challenging problem.

Product Rule

The Product Rule is crucial for functions that are products of two or more parts. It's used when differentiating a function that multiplies two components. The formula is:If you have two functions, say \(u(x)\) and \(v(x)\), then the derivative is:\[f'(x) = u'(x)v(x) + u(x)v'(x)\]In the exercise given, we have the function \(f(x) = xe^x - e^x\). Notice that first, we treat \(xe^x\) as a single function where the Product Rule is applied. It means taking the derivative of the first function \(x\), which is 1, and multiplying it by the second function \(e^x\), then adding the product of the first function \(x\) and the derivative of \(e^x\). The derivative is calculated as:- First derivative: \(1 \cdot e^x + x \cdot e^x = xe^x\).

Second Derivative Test

The Second Derivative Test helps determine whether a critical point is a local maximum, a local minimum, or neither. This test involves taking the second derivative of the function.When you find a critical point \(x = c\):

• If \(f''(c) > 0\), the function has a local minimum at \(x = c\).
• If \(f''(c) < 0\), there is a local maximum at \(x = c\).
• If \(f''(c) = 0\), the test is inconclusive.

In our exercise, after computing the second derivative \(f''(x) = (2 + x)e^x\), we evaluated it at the critical point \(x = 0\):- Calculating \(f''(0)\) gives 2, indicating a local minimum.

Local Maxima and Minima

Local maxima and minima are points where a function changes direction. It means the function has a peak (maximum) or a trough (minimum).To find local maxima and minima, follow these steps:

• Compute the first derivative and identify critical points where \(f'(x) = 0\) or is undefined.
• Use the second derivative test to determine the nature of these points.

In our exercise, after finding that the critical point is \(x = 0\) with the second derivative value \(f''(0) = 2\), we concluded that our function has a local minimum at this point because \(f''(0) > 0\). This helps to sketch the graph and understand the function's behavior in that particular region.