Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find, if possible, the (global) maximum and minimum values of the given function on the indicated interval. $$ f(x)=x e^{-x} \text { on }[0, \infty) $$

Step-by-Step Solution

Answer

The maximum value is \( e^{-1} \approx 0.368 \) at \( x = 1 \), and the minimum is \( 0 \) at \( x = 0 \).

Step 1: Identify the Function and Interval

The function given is \( f(x) = x e^{-x} \) and the interval we are considering is \( [0, \infty) \). Our task is to find if there are global maximum and minimum values on this interval.

Step 2: Find the Derivative

Calculate the derivative of \( f(x) \). We use the product rule for differentiation. Given \( u = x \) and \( v = e^{-x} \), their derivatives are \( u' = 1 \) and \( v' = -e^{-x} \). Thus the derivative is:\[ f'(x) = (x) \cdot (-e^{-x}) + (1) \cdot (e^{-x}) = e^{-x} (1 - x) \]

Step 3: Find Critical Points

Set the derivative \( f'(x) = e^{-x} (1 - x) \) equal to zero to find critical points. Because \( e^{-x} eq 0 \) for all \( x \), we solve: \[ 1 - x = 0 \]Thus, \( x = 1 \) is a critical point.

Step 4: Evaluate Function at Critical Points and Endpoints

Evaluate \( f(x) \) at the critical point and at the endpoints if applicable:- \( f(0) = 0 \cdot e^{0} = 0 \)- \( f(1) = 1 \cdot e^{-1} \approx 0.368 \)- As \( x \to \infty \), \( f(x) = x e^{-x} \to 0 \) because the exponential term \( e^{-x} \) dominates.Compare these values to find the maximum and minimum.

Step 5: Determine Maximum and Minimum Values

From the evaluations:- The minimum value is \( f(0) = 0 \).- The maximum value is \( f(1) = e^{-1} \approx 0.368 \).

Key Concepts

Derivative Calculation

Calculus optimization often begins with calculating the derivative of a function. This single step provides insight into how the function changes. In our exercise, we need to find the derivative of the function \( f(x) = x e^{-x} \).
Using the product rule is crucial here, as the function is a product of two terms: \( u = x \) and \( v = e^{-x} \).

• The derivative of \( u = x \) is \( u' = 1 \).
• The derivative of \( v = e^{-x} \) is \( v' = -e^{-x} \).

We apply the product rule for differentiation, which states that the derivative of a product \( uv \) is \( u'v + uv' \). Thus, the derivative \( f'(x) \) becomes \[ f'(x) = (1)(e^{-x}) + (x)(-e^{-x}) = e^{-x}(1-x). \]
This derivative is a fundamental building block for finding critical points and analyzing the behavior of the original function.

Critical Points

Critical points of a function are where its derivative equals zero or is undefined. These points are key to finding local and global extrema. For \( f(x) = x e^{-x} \), we solve \( f'(x) = e^{-x} (1 - x) = 0 \) to locate critical points.
Since the exponential component \( e^{-x} \) is never zero for any real \( x \), we focus on solving \( 1 - x = 0 \). This leads to the critical point at \( x = 1 \). This location is where the slope of \( f(x) \) changes from increasing to decreasing, or vice versa, and needs to be tested for extrema.
Critical points like this are where we check if the function reaches a local peak or valley.

Global Maximum and Minimum

Once critical points are identified, we evaluate the function at these points, as well as at the endpoints of the interval, to determine the global maximum and minimum.

• For \( f(x) \) on \([0, \infty)\), the value at the critical point \( x = 1 \) is \( f(1) = e^{-1} \approx 0.368 \).
• The endpoint \( x = 0 \) yields \( f(0) = 0 \).

As \( x \) approaches infinity, \( f(x) = x e^{-x} \rightarrow 0 \) because the decay of \( e^{-x} \) is faster than the growth of \( x \).
The analysis shows that the maximum function value on the interval is \( f(1) = e^{-1} \), and the minimum is at \( f(0) = 0 \). Understanding where a function peaks and bottoms out helps find these essential values.

Interval Analysis

Analyzing the interval is the final step to confirming the global extremum values. With \( f(x) = x e^{-x} \) on \([0, \infty)\), we're tasked with observing behaviors at specific points.

• Critical point evaluations reveal behavior changes at those points, like \( x = 1 \).
• Endpoints and approaches to infinity provide bounds for function values.

This approach helps ensure that we do not miss any potential extremum outside our calculated critical points.
In our case, noting that as \( x \to \infty \), the function \( f(x) = x e^{-x} \) tends to zero, reaffirms that \( x = 0 \) and \( x = 1 \) indeed cover all maximum and minimum concerns.
By thoroughly dissecting the interval, either through intuitive reasoning or calculated evaluation, one can confidently state where a function peaks or troughs, crucial for solving optimization problems.