Problem 45 - 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

As \(x \rightarrow +\infty\), \(f(x) \rightarrow +\infty\). As \(x \rightarrow -\infty\), \(f(x) \rightarrow 0\). Relative minimum at \(x = -1\), inflection point at \(x = -2\).

Step 1: Identify the Limits

First, identify the limits of the function as given in the problem for large positive and negative values of \(x\). For \(f(x) = xe^x\), calculate the limit as \(x\) approaches infinity and negative infinity. We already know from the exercise statement that \(\lim_{x \rightarrow +\infty} \frac{x}{e^x} = 0\) and \(\lim_{x \rightarrow -\infty} x \cdot e^x = 0\). Thus, \(\lim_{x \rightarrow +\infty} xe^x = +\infty\) and \(\lim_{x \rightarrow -\infty} xe^x = 0\).

Step 2: Differentiate to Find Extrema and Inflection Points

Find the first derivative \(f'(x)\) to determine relative extrema, using \(f(x) = xe^x\). The derivative using the product rule is \(f'(x) = e^x + xe^x\). Set \(f'(x) = 0\) to find critical points. Simplifying, we solve \(e^x(1 + x) = 0\), which leads to \(x = -1\) as the critical point. Find the second derivative, \(f''(x) = 2e^x + xe^x\), setting \(f''(x) = 0\) to find inflection points. This gives \(x = -2\) as the point of inflection.

Step 3: Determine Asymptotes

The function \(f(x) = xe^x\) does not have any horizontal or vertical asymptotes. As \(x\) approaches positive infinity, \(f(x)\) approaches positive infinity due to the dominance of \(e^x\). As \(x\) approaches negative infinity, \(f(x)\) approaches zero.

Step 4: Sketch the Graph

Draw a rough sketch of \(f(x) = xe^x\) based on the information gathered. The graph should show the function approaching zero as \(x\) goes to negative infinity, a minimum at \(x = -1\), an inflection point at \(x = -2\), and increasing rapidly to infinity as \(x\) goes to positive infinity. Use a graphing utility to verify the sketch.

Key Concepts

Limits of Exponential Functions

Exponential functions are unique because they grow very rapidly as the input increases. When we talk about the limits of these functions, we focus on their behavior as the variable, in this case, \(x\), approaches positive or negative infinity.

In the provided exercise, we encounter limits such as \(\lim_{x \rightarrow +\infty} \frac{x}{e^x} = 0\) and \(\lim_{x \rightarrow -\infty} x \cdot e^x = 0\).

• As \(x\) approaches positive infinity, \(e^x\) grows much faster than any polynomial term like \(x\). This means that \(\frac{x}{e^x} \rightarrow 0\).
• As \(x\) approaches negative infinity, \(e^x\) becomes very small (approaching zero), which means the product \(x \cdot e^x\) also tends towards zero.

Understanding these behaviors is crucial because they dictate the overall outcome of larger expressions such as \(x \cdot e^x\) when \(x\) moves towards infinity in both directions.

Critical Points

Critical points are values of \(x\) where the first derivative of a function is zero or undefined. Identifying these points is important because they indicate potential locations of local maxima, minima, or points where the function's increasing or decreasing behavior changes.

For the function \(f(x) = xe^x\), we calculated the derivative \(f'(x) = e^x(1 + x)\) and set it equal to zero to find critical points. Solving \(e^x(1+x) = 0\) gives us \(x = -1\) because \(e^x\) is never zero, and thus \(1 + x\) must be zero.

• This critical point \(x = -1\) suggests a relative extremum, specifically a minimum, due to how the function behaves around this point.

Understanding critical points helps determine where the function changes direction, which is essential for sketching and analyzing the behavior of the graph.

Derivative Tests

Derivative tests use the first and second derivatives to determine more about a function's potential relative extrema and inflection points.

From the first derivative \(f'(x) = e^x + xe^x\), setting this to zero helped identify the critical point at \(x = -1\). After determining this point, we use the second derivative test with \(f''(x) = 2e^x + xe^x\) to check the nature of this critical point.

• If the second derivative at \(x = -1\) is positive, it's a local minimum.
• If it's negative, it's a local maximum. In this instance, \(f''(-1) > 0\), confirming \(x = -1\) is a minimum.

By further setting \(f''(x) = 0\), we find inflection points where the concavity of the graph changes:\(x = -2\) in this case.
The derivative tests provide insight into the graph's shape and behavior, helping us understand where it bends and turns.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as the input approaches infinity or some boundary of its domain.

For \(f(x) = xe^x\), we look at limits as \(x\) approaches both positive and negative infinity to understand the asymptotic behavior.

• As \(x \to +\infty\), \(f(x)\) grows rapidly towards positive infinity with no horizontal or vertical asymptotes because the exponential growth dominates.
• As \(x \to -\infty\), \(f(x)\) approaches zero due to the diminishing value of \(e^x\), giving a horizontal asymptote at \(y = 0\).

In this function, we don't have vertical asymptotes because it's defined for all real numbers. Understanding asymptotic behavior helps us predict long-term trends of the function and ensures we accurately represent its behavior as the graph extends far in either direction.