Problem 93 - Calculus Single Variable

Question

In each of Exercises \(91-94\), calculate and plot the derivative \(f^{\prime}\) of the given function \(f\). Use this plot to identify candidates for the local extrema of \(f\). Add the plot of \(f\) to the window containing the graph of \(f^{\prime} .\) From this second plot, determine the behavior of \(f\) at each candidate for a local extremum. $$ f(x)=x-2 \exp \left(-x^{2}\right) $$

Step-by-Step Solution

Answer

Find the derivative \(f'(x) = 1 + 4xe^{-x^2}\); use the plot to identify local extrema candidates and analyze \(f(x)\) behavior at these points.

Step 1: Find the Derivative

First, differentiate the given function \( f(x) = x - 2e^{-x^2} \). This will involve using the chain rule for the derivative of an exponential function.The derivative of the function \( f(x) \) is calculated as follows:\[ f'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(2e^{-x^2}) \]The derivative of \( x \) is 1, and the derivative of \( 2e^{-x^2} \) involves the chain rule:\[ \frac{d}{dx}(2e^{-x^2}) = 2 \times -2xe^{-x^2} = -4xe^{-x^2} \]Thus, the derivative of the function is:\[ f'(x) = 1 + 4xe^{-x^2} \]

Step 2: Plot the Derivative

Plot the derivative \( f'(x) = 1 + 4xe^{-x^2} \). This plot will help us to identify the points where the derivative changes sign, indicating possible local extrema.

Step 3: Identify Candidates for Local Extrema

From the plot of \( f'(x) \), find points where the derivative crosses the x-axis, which correspond to points where \( f'(x) = 0 \). These are candidates for local extrema of the original function \( f(x) \). Analyze whether the plot suggests any local maximum or minimum.

Step 4: Plot the Original Function

Now plot the original function \( f(x) = x - 2e^{-x^2} \) in the same window as the plot of \( f'(x) \). This allows us to visually confirm the behavior of \( f(x) \) at the candidate extrema identified from the derivative plot.

Step 5: Determine the Behavior at Local Extrema

Analyze the plot of \( f(x) \) overlaid with the plot of \( f'(x) \). Look at the behavior of the function \( f(x) \) at each candidate for a local extremum identified in Step 3. If the original function has a slope changing from positive to negative at a point, it is a local maximum. If the slope changes from negative to positive, it is a local minimum.

Key Concepts

Chain Rule

The Chain Rule is an essential concept in calculus, particularly when dealing with complex functions that involve an inner function and an outer function. In the exercise, we encounter the function:

• \[ f(x) = x - 2 \exp(-x^2) \]

The task is to differentiate this function. Notice the component \(-2 \exp(-x^2)\), which requires the Chain Rule because it involves a composite function. The inner function here is \( -x^2 \) and the outer function is the exponential \( e^u \), where \( u\) is the inner function.
Thus, we differentiate the inner function and multiply it by the derivative of the outer function. We start with:

• The derivative of \( x \) is simply \( 1 \).
• The derivative of \( -x^2 \) is \( -2x \).
• Then, multiply this by the derivative of the outer function \( e^{-x^2} \), which remains \( e^{-x^2} \), giving \( -2x \cdot e^{-x^2} \).

Therefore, the derivative becomes:

• \[ f'(x) = 1 + 4x \exp(-x^2) \]

This demonstrates how the Chain Rule helps in tackling derivatives of composite functions, making it a powerful tool in calculus.

Local Extrema

Local extrema refer to places on the curve of a function where the function reaches a maximum or minimum value relative to its immediate vicinity. In our given function, we seek points where the derivative changes sign, indicating potential local maxima or minima.To identify these, consider that such points occur when the derivative \( f'(x) \) set to zero:

• The equation \( 1 + 4x \exp(-x^2) = 0 \) needs to be solved for \( x \).

Finding when the derivative equals zero gives:

• \[ 4x \exp(-x^2) = -1 \], implying potential extrema.

Once you identify these points, it is essential to determine the nature of each extremum:

• If the derivative moves from positive to negative, the point is a local maximum.
• Conversely, if it changes from negative to positive, you have a local minimum.

This process is significant because it explains the behavior of a function around critical points and gives a deeper understanding of its structure.

Derivative Plot

Plotting the derivative provides a visual representation of how the rate of change of the function \( f(x) \) behaves over its domain. It helps you quickly spot the zero crossings, where the function might have local extrema.In this exercise, once we have the derivative:

• \[ f'(x) = 1 + 4x \exp(-x^2) \]

The next step is to plot this derivative.
The plot can reveal places where the curve crosses the x-axis, indicating potential maxima or minima. For this function, plotting also helps understand:

• The behavior of the slope around critical points.
• Where \( f(x) \) either rises or falls, indicating changes in the original function slope.

By overlaying this derivative plot with the plot of the original function, one can connect the calculated mathematical behavior with actual visual interpretation.
This combination offers a comprehensive toolset for identifying and understanding the intricacies of function behavior at or around critical points. It also makes the abstract concept of differentiation more tangible and applicable, easing the learning process.