Problem 104 - Calculus and its applications

Question

Find the minimum value of \(f(x)=x e^{x}\) over [-2,0]

Step-by-Step Solution

Answer

The minimum value of \( f(x) = x e^x \) over \([-2, 0]\) is \(-\frac{2}{e^2}\).

Step 1: Understand the Function

To find the minimum value of the function \( f(x) = x e^{x} \) over the interval \([-2, 0]\), we need to evaluate where the function increases or decreases by finding its derivative.

Step 2: Find the Derivative

Differentiate the function \( f(x) = x e^x \) using the product rule: if \( u = x \) and \( v = e^x \), then \( u' = 1 \) and \( v' = e^x \). Therefore, the derivative is: \[f'(x) = e^x + x e^x = e^x(1 + x)\]

Step 3: Find the Critical Points

Set the derivative equal to zero to find the critical points: \[ e^x (1 + x) = 0 \].Since \( e^x \) is never zero, solve \( 1 + x = 0 \), giving \( x = -1 \).

Step 4: Evaluate at Critical Points and Endpoints

Evaluate \( f(x) \) at the critical point \( x = -1 \) and at the endpoints \( x = -2 \) and \( x = 0 \). \[ f(-2) = (-2)e^{-2} = -\frac{2}{e^2} \]\[ f(-1) = (-1)e^{-1} = -\frac{1}{e} \]\[ f(0) = 0 \, e^0 = 0 \].

Step 5: Determine the Minimum Value

Compare the values from the previous step:- \( f(-2) = -\frac{2}{e^2} \)- \( f(-1) = -\frac{1}{e} \)- \( f(0) = 0 \)The smallest value is \( f(-2) \), thus the minimum value of the function over the interval \([-2, 0]\) is \(-\frac{2}{e^2}\).

Key Concepts

Derivative Calculations

Finding the derivative of a function is a fundamental task in calculus. It helps us understand how a function behaves by determining where it increases or decreases. The derivative of a function at any given point represents the slope of the tangent line to the graph at that point. For the function \( f(x) = x e^x \), we use differentiation to find the rate of change.Remember, the first step is to identify the function parts to be differentiated. Here we have:

• \( u = x \)
• \( v = e^x \)

Differentiating these individually gives us \( u' = 1 \) and \( v' = e^x \). We then apply the product rule to find the overall derivative. Understanding these foundational tasks makes finding critical points and analyzing functions much easier.

Critical Points Analysis

Critical points of a function are where its derivative is zero or undefined. These points are significant because they are potential locations for local maxima, minima, or points of inflection. In the problem, the critical points of the function \( f(x) = x e^x \) were found by solving the equation:\[ f'(x) = e^x (1 + x) = 0 \]Since \( e^x \) is always positive (never zero), we solve for \( x \) when \( 1 + x = 0 \). Solving this gives us \( x = -1 \). It’s essential to remember that finding where the derivative equals zero is just the start. Further investigation, like using the actual function values at these points, helps in determining whether these points are indeed minima or maxima.

Function Evaluation

Evaluating a function involves finding the actual output values at specific points of interest, such as endpoints and critical points. This step allows us to draw conclusions about the function's behavior. In our case, we calculated \( f(x) \) for \( x = -2, -1, \text{and } 0 \). Each of these points gave us:

• \( f(-2) = -\frac{2}{e^2} \)
• \( f(-1) = -\frac{1}{e} \)
• \( f(0) = 0 \)

Comparing these values helps to determine the absolute minimum of the function within the interval \( [-2, 0] \). It’s a crucial procedure to validate the results obtained from derivative analysis.

Product Rule in Calculus

The product rule is a key technique in calculus used specifically for finding the derivative of the product of two functions. When one function is dependent on another, as in \( f(x) = x e^x \), the rate of change is affected by both parts.The formula for the product rule is: \[ (uv)' = u'v + uv' \]Applying it to our function where \( u = x \) and \( v = e^x \), you differentiate each part and then substitute back:\[ f'(x) = 1 \cdot e^x + x \cdot e^x = e^x + x e^x \]Thus, the derivative becomes \( e^x (1 + x) \). This illustrates the utility of the product rule in simplifying complex differentiation tasks. Mastering this rule opens up the ability to tackle a wide variety of more complex functions in calculus.