Problem 16 - Calculus Single Variable

Question

In each of Exercises \(7-22,\) use Fermat's Theorem to locate each \(c\) for which \(f(c)\) is a candidate extreme value of the given function \(f\) $$ f(x)=x e^{x} $$

Step-by-Step Solution

Answer

The candidate extreme value is at \( c = -1 \).

Step 1: Write down Fermat's Theorem

Fermat's Theorem states that if a function has a local extremum at a point, and if the derivative exists at that point, then the derivative is zero. In mathematical terms, if \( f'(c) = 0 \), then \( c \) is a candidate for a local extremum.

Step 2: Find the Derivative of the Function

First, we need to find the derivative of the given function \( f(x) = x e^x \). Use the product rule for derivatives, which states \( (uv)' = u'v + uv' \). Here, \( u = x \) and \( v = e^x \).

Step 3: Apply the Product Rule

Find the derivatives \( u' \) and \( v' \). We have \( u = x \), so \( u' = 1 \), and \( v = e^x \), so \( v' = e^x \). According to the product rule, \( f'(x) = 1 \cdot e^x + x \cdot e^x = e^x + xe^x \).

Step 4: Set the Derivative Equal to Zero

Now, set the derivative to zero to find critical points: \( e^x + xe^x = 0 \). Factor out \( e^x \) to get: \( e^x(1 + x) = 0 \).

Step 5: Solve for Critical Points

Since \( e^x \) is never zero for any real number \( x \), we only need to consider \( 1 + x = 0 \). Solve for \( x \) to find \( x = -1 \). Thus, \( c = -1 \) is a candidate for an extreme value.

Key Concepts

Critical Points

A critical point of a function is where its derivative is zero or undefined. This point is significant as it potentially indicates where the function has a local max or min, though not all critical points are extrema.

Critical points are identified by examining the derivative of the function. For example, if a function is differentiable at a certain point and its derivative at that point is zero, it qualifies as a critical point. These points are essential in optimization and understanding the behavior of functions.

• Step 1: Find the first derivative of the function.
• Step 2: Set the derivative equal to zero or determine where it does not exist.
• Step 3: Solve for the values of x, these solutions are the critical points.

Product Rule

When dealing with derivatives of products of functions, the product rule is a key tool. It helps in finding derivatives of expressions where two functions are multiplied.
For two functions, say \( u(x) \) and \( v(x) \), the product rule is defined as:
\[ (uv)' = u'v + uv' \]
In the given exercise, the product rule is applied to the function \( f(x) = x e^x \). Here you identify \( u = x \) and \( v = e^x \). Calculating their derivatives, we find \( u' = 1 \) and \( v' = e^x \). Applying the product rule gives us the derivative: \( f'(x) = e^x + x e^x \).

This method is essential for analyzing composite functions and is frequently used in calculus for finding derivatives effectively.

Derivative

A derivative gives the rate at which a function is changing at any point. It serves as the building block for calculus, allowing us to understand functions' local behavior.
To find the derivative of any function, apply the rules of differentiation, such as the power rule, product rule, and chain rule, when necessary.

Use the derivative to identify where a function increases or decreases, and find critical points by setting it to zero. In the function \( f(x) = x e^x \), the derivative is found using the product rule: \( f'(x) = e^x + x e^x \).

• The derivative helps locate critical points, essential for finding extrema in functions.
• A key concept in calculus, derivatives provide insight into the graph of a function.

Local Extremum

A local extremum of a function is a point where the function reaches a local maximum or minimum. At these points, the function "turns around" in the graph.
A local minimum is where a function's value is lower than the values around it, and a local maximum is where the function's value is higher than its surrounding values. Mathematical determination involves finding where the function's derivative is zero and using tests like the second-derivative test to classify these points.

Fermat's theorem plays a pivotal role here, stating if a function has a local extremum at point \( c \), and the derivative exists there, then \( f'(c) = 0 \). In our example, the critical point \( x = -1 \) is a candidate for local extremum, as obtained by setting \( e^x(1 + x) = 0 \).

Understanding local extrema aids in analyzing function graphs, optimizing problems, and solving real-world issues effectively.