Problem 45 - Calculus A New Horizon

Question

(a) Find the limits of the function as \(x \rightarrow 0^{+}\) and \(x \rightarrow+\infty .\) (b) Give a complete graph of the function, and identify the location of all relative extrema and inflection points. Check your work with a graphing utility. $$x \ln x$$

Step-by-Step Solution

Answer

As \( x \to 0^+ \), \( x \ln x \to 0 \). As \( x \to \infty \), \( x \ln x \to \infty \). The function has a relative minimum at \( x = e^{-1} \) and no inflection points.

Step 1: Find the Limit as x Approaches 0 from the Right

To find \( \lim_{{x \to 0^+}} x \ln x \), we recognize that both \( x \rightarrow 0^+ \) and \( \ln x \rightarrow -\infty \). Therefore, this is a \( 0 \times -\infty \) indeterminate form. Rewriting \( x \ln x \) as \( \frac{\ln x}{1/x} \) allows us to use L'Hospital's Rule. Differentiating the numerator and the denominator gives us:\[ \lim_{{x \to 0^+}} \frac{\ln x}{1/x} = \lim_{{x \to 0^+}} \frac{1/x}{-1/x^2} = \lim_{{x \to 0^+}} -x = 0. \]Thus, \( \lim_{{x \to 0^+}} x \ln x = 0. \)

Step 2: Find the Limit as x Approaches Infinity

To find \( \lim_{{x \to \infty}} x \ln x \), note that \( x \rightarrow \infty \) and \( \ln x \rightarrow \infty \), making both components positive and infinite, thus:\[ \lim_{{x \to \infty}} x \ln x = \infty. \]Because both functions tend towards their positive infinities, the limit is \( \infty \).

Step 3: Find Critical Points for Relative Extrema

The critical points for function \( y = x \ln x \) occur when \( y' = 0 \). First, find the derivative using the product rule:\[ y' = \ln x + 1. \]Set \( y' = 0 \):\[ \ln x + 1 = 0 \quad \Rightarrow \quad \ln x = -1 \quad \Rightarrow \quad x = e^{-1}. \]This is where a critical point may occur.

Step 4: Determine Relative Extrema Using Second Derivative

Find \( y'' \):\[ y'' = \frac{d}{dx}(\ln x + 1) = \frac{1}{x}. \]Evaluate at \( x = e^{-1} \):\[ y''(e^{-1}) = \frac{1}{e^{-1}} = e > 0. \]Since \( y'' > 0 \), the function has a relative minimum at \( x = e^{-1} \).

Step 5: Identify Inflection Points

An inflection point occurs where \( y'' = 0 \) or changes sign, but since \( y'' = \frac{1}{x} \) is never zero and does not change sign, there are no inflection points.

Step 6: Sketch the Graph

The function \( y = x \ln x \) has a critical point at \( x = e^{-1} \) (a relative minimum). The limits at extreme values are determined: approaching 0 from the right yields 0, and approaching infinity yields infinity. The graph starts from zero near \( x = 0 \), descends to a local minimum at \( x = e^{-1} \), then increases without bound.

Key Concepts

L'Hospital's Rule

Sometimes in calculus, we encounter indeterminate forms when trying to find limits. Indeterminate forms can be perplexing because they do not intuitively suggest a limit. This is where L'Hospital's Rule comes to the rescue. It provides a systematic way to resolve such forms, specifically leading to

• 0/0
• ∞/∞

forms by differentiating the numerator and the denominator.
In the given exercise, the function when considered as \( x \ln x \) becomes \( \frac{\ln x}{1/x} \), which is a 0 times -∞ indeterminate form. By differentiating the numerator and denominator separately, we receive a simplified function. Eventually, it leads us to
\[ \lim_{{x \to 0^+}} \frac{1/x}{-1/x^2} \]
and evaluating this expression gives us a neat solution of 0.

Relative Extrema

Determining the peaks and valleys in a function's graph helps identify key turning points known as relative extrema. The first derivative test helps us locate these points by solving for when the derivative equals zero.
For the function \( y = x \ln x \), its derivative using the product rule simplifies to
\[ y' = \ln x + 1. \]Finding critical points involves solving
\[ \ln x + 1 = 0 \Rightarrow x = e^{-1}. \]
This critical point tells us where potential extrema lie. Using the second derivative, \( y'' = \frac{1}{x} \), we can analyze the nature of these extremum points. Since \( y''(e^{-1}) > 0 \), the function has a relative minimum at this point. It's a spot where the function transitions from decreasing to increasing behavior.

Inflection Points

Inflection points are key in analyzing the concavity of functions. They occur when a function switches between being concave up and concave down, typically noted where the second derivative equals zero or changes sign.
However, in our case of \( y = x \ln x \), analyzing the second derivative
\[ y'' = \frac{1}{x} \]
reveals no zeros due to its non-zero value for all \( x > 0 \). Given that it doesn't change sign either, our function lacks inflection points.
This information indicates that the curvature of the function is consistent and doesn't flip, contributing to our understanding of the graph's behavior.

Product Rule

The product rule is essential to differentiating expressions where two functions are multiplied. When dealing with products of functions, it applies efficiently to finding derivatives.

• Given functions \( u(x) \) and \( v(x) \), the product rule states:
• \( (uv)' = u'v + uv'. \)

In the scenario of the function \( y = x \ln x \), seeing this as a product of \( x \) and \( \ln x \) means it is necessary to apply the rule to find:
\[ y' = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1. \]
This step is vital in identifying critical points, revealing behavior changes within the function. By comprehensively understanding the product rule, differentiating more complex expressions becomes manageable and clear.