Problem 46 - Calculus Early Transcendentals

Question

Using L'Hôpital's rule (Section 3.6) one can verify that $$ \lim _{x \rightarrow+\infty} \frac{e^{x}}{x}=+\infty, \quad \lim _{x \rightarrow+\infty} \frac{x}{e^{x}}=0, \quad \lim _{x \rightarrow-\infty} x e^{x}=0 $$ In these exercises: (a) Use these results, as necessary, to find the limits of \(f(x)\) as \(x \rightarrow+\infty\) and as \(x \rightarrow-\infty\). (b) Sketch a graph of \(f(x)\) and identify all relative extrema, inflection points, and asymptotes (as appropriate). Check your work with a graphing utility. $$ f(x)=x e^{-x} $$

Step-by-Step Solution

Answer

Both limits of \( f(x) = x e^{-x} \) as \( x \rightarrow \pm \infty \) are 0. The function peaks at \( x=1 \).

Step 1: Analyze Limit as x Approaches Infinity

We need to find \( \lim _{x \rightarrow+\infty} f(x) = \lim _{x \rightarrow+\infty} x e^{-x} \). Rewrite the expression as \( \frac{x}{e^x} \). From the given results, we know \( \lim _{x \rightarrow+\infty} \frac{x}{e^x} = 0 \). Therefore, \( \lim _{x \rightarrow+\infty} f(x) = 0 \).

Step 2: Analyze Limit as x Approaches Negative Infinity

Evaluate \( \lim _{x \rightarrow-\infty} f(x) = \lim _{x \rightarrow-\infty} x e^{-x} \). Rewrite the expression as \( \lim _{x \rightarrow-\infty} \frac{x}{e^x} \). Given that \( \lim _{x \rightarrow-\infty} x e^x = 0 \), we can conclude \( \lim _{x \rightarrow-\infty} f(x) = 0 \).

Step 3: Sketch the Graph of f(x) and Identify Extrema and Inflection Points

To sketch \( f(x) = x e^{-x} \), first find critical points by setting the derivative \( f'(x) = e^{-x} - x e^{-x} \) to zero. This simplifies to \( 1 - x = 0 \), giving the critical point \( x = 1 \). Evaluate \( f''(x) = -e^{-x} + x e^{-x} \) to find inflection points. Setting \( f''(x) = 0 \), we solve \( -1 + x = 0 \), finding \( x = 1 \) as a potential inflection point as well. Check concavity around it and verify with a graphing utility. The curve decreases as \( x \rightarrow -\infty \) and initially increases to \( x = 1 \) before decreasing again as \( x \rightarrow \infty \). No asymptotes are present other than the behavior as \( x \rightarrow \pm \infty \).

Step 4: Confirm and Reflect with Graphing Tool

Use a graphing utility to verify the sketch. Ensure the behavior of \( f(x) \) matches the expected limits and critical behavior, with the function approaching zero as \( x \rightarrow \pm \infty \), achieving a peak at \( x = 1 \). This confirms the presence of a relative maximum and the inflection point at \( x = 1 \). Retracing the sketched graph with the tool may reveal consistent concavity changes and confirm the smooth transition of values.

Key Concepts

Understanding Limits and Their Application with L'Hôpital's Rule

Limits help us understand the behavior of functions as they approach certain values. When facing an indeterminate form, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hôpital's Rule becomes helpful. This rule states that if the limit \(\lim_{x \to c} \frac{f(x)}{g(x)}\) is in an indeterminate form, then it can be evaluated as \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\), provided this new limit exists. For example, to find \(\lim_{x \to +\infty} \frac{x}{e^x}\) in our problem, L'Hôpital's Rule simplifies the process since it initially forms \(\frac{\infty}{\infty}\). Differentiating the numerator and the denominator, we get \(\frac{1}{e^x}\), simplifying the evaluation. This makes the limit approach zero, confirming \(f(x)\) approaches zero as \(x\) goes to infinity.
The same approach is used when \(x\) approaches negative infinity. Exploring these limits is crucial for understanding the function's behavior, especially near infinity, which is essential in sketching graphs and assessing asymptotic behavior.

Identifying Relative Extrema in Functions

Relative extrema refer to points on a graph where the function reaches a local maximum or minimum. To find these, we examine where the derivative of the function \(f'(x)\) is zero or undefined. In our exercise, the function \(f(x) = xe^{-x}\) has a derivative \(f'(x) = e^{-x} - xe^{-x}\). By simplifying and setting this to zero, we find the critical point at \(x=1\).
Evaluating \(f''(x)\) at these critical points helps determine their nature - whether they are maxima, minima, or inflection points. In our case, analyzing \(f(x)\) reveals that \(x = 1\) is a point where the function reaches a relative maximum. Understanding how the derivative changes allows us to visualize the peaks and valleys of a graph, guiding us in sketching functions accurately.

Understanding Inflection Points in Graphs

Inflection points occur where a function changes the direction of its curvature. They are found where the second derivative \(f''(x)\) changes sign. Analyzing \(f''(x) = -e^{-x} + xe^{-x}\) for the function \(f(x) = xe^{-x}\), setting \(f''(x) = 0\) helps identify potential changes in concavity. Solving this gives us \(x = 1\) as a candidate for an inflection point.
Inflection points play a vital role in understanding the overall shape of a graph. They tell us about areas where the function transitions from concave up to concave down or vice versa. Knowing the presence and location of inflection points aids in sketching the graph accurately and understanding its characteristics. Inflection points, combined with relative extrema and limits, provide a comprehensive picture of the function's behavior.