Problem 14 - Applied Calculus

Question

Describe in words the interesting features of the graph, including the location of the critical points and where the function is monotonic (that is, increasing or decreasing). Then use the derivative and algebra to explain the shape of the graph. $$ f(x)=x \ln x, \quad x>0 $$

Step-by-Step Solution

Answer

The function has a local minimum at \(x = \frac{1}{e}\) and decreases on \((0, \frac{1}{e})\), increasing on \(x > \frac{1}{e}\).

Step 1: Identify the Domain of the Function

The function given is \(f(x) = x \ln x\) and is only defined for \(x > 0\) because the natural logarithm, \(\ln x\), is undefined for \(x \leq 0\). Thus, the domain of the function is \(x > 0\).

Step 2: Find the First Derivative

To understand the critical points and monotonicity, we first find the derivative of \(f(x) = x \ln x\). Using the product rule, which states that \( (uv)' = u'v + uv' \), let \( u = x \) and \( v = \ln x \):\[ f'(x) = (x)'(\ln x) + x(\ln x)' = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1. \]

Step 3: Find Critical Points

Critical points occur where the derivative \( f'(x) = \ln x + 1 \) is equal to zero or undefined in its domain. Setting \(f'(x) = 0\) gives:\[\ln x + 1 = 0\]Solving for \(x\) yields \( \ln x = -1 \), which means \(x = e^{-1} = \frac{1}{e}\). Thus, there is a critical point at \(x = \frac{1}{e}\).

Step 4: Determine Monotonicity

To determine where \(f(x)\) is increasing or decreasing, examine \(f'(x) = \ln x + 1\). - For \(x > \frac{1}{e}\), \(\ln x > -1\), thus \(f'(x) > 0\) and the function is increasing.- For \(0 < x < \frac{1}{e}\), \(\ln x < -1\), thus \(f'(x) < 0\) and the function is decreasing.

Step 5: Summarize Interesting Features

The graph of \(f(x) = x \ln x\) has a critical point at \(x = \frac{1}{e}\), which is a local minimum because the function transitions from decreasing to increasing at this point. The graph decreases from \((0, \frac{1}{e})\), is at its minimum at \(x = \frac{1}{e}\), and increases for \(x > \frac{1}{e}\).

Key Concepts

Derivative

In calculus, a derivative measures how a function changes as its input changes. It tells us the slope of the tangent line to the graph of the function at any given point. This is crucial for understanding the behavior of functions. For the function given, \( f(x) = x \ln x \), its derivative helps us learn about the graph’s shape. We find the derivative to spot crucial features, like where the function climbs or descends. Derivatives let us diagnose the character of the function at each point, paving the path to deeper insights about the graph’s behavior.
To find the derivative of \( f(x) = x \ln x \), we used the product rule, an essential tool for functions made from multiplying two expressions. This is nicely detailed in the following section.

Critical Points

Critical points are where the derivative is zero or undefined. They highlight important changes in the function's behavior. For the function \( f(x) = x \ln x \), the critical point is where \( f'(x) = \ln x + 1 = 0 \). Solving this, we get \( x = \frac{1}{e} \).
Critical points are not just numbers; they tell us where the function reaches peaks (maxima), valleys (minima), or points of inflection. In this problem, there is a local minimum at \( x = \frac{1}{e} \). The graph transitions from a descending segment to an ascending one here, offering us a vital piece about the function’s nature.

Monotonicity

Monotonicity refers to the intervals where a function is consistently increasing or decreasing. Understanding a function's monotonicity helps us predict its overall shape. For our function \( f(x) = x \ln x \), the derivative \( f'(x) = \ln x + 1 \) enables this insight.
For \( x > \frac{1}{e} \), we find \( f'(x) > 0 \); thus, the function is increasing here. In the interval \( 0 < x < \frac{1}{e} \), where \( f'(x) < 0 \), the function is decreasing. This breakdown tells us the function drops to a minimum at \( x = \frac{1}{e} \) and then rises. Such insights allow us to sketch the big picture of the graph’s dynamics.

Product Rule

The product rule is a fundamental technique in calculus used to differentiate functions formed by multiplying two other functions. It states that if you have a function \( h(x) \) defined as \( h(x) = u(x) \cdot v(x) \), then its derivative is \( h'(x) = u'(x) v(x) + u(x) v'(x) \).
For our function \( f(x) = x \ln x \), let’s consider \( u = x \) and \( v = \ln x \). Applying the product rule, we differentiate each part separately and combine the results: \( f'(x) = (x)'(\ln x) + x(\ln x)' = \ln x + 1 \).
This rule is powerful, allowing us to handle more complex functions efficiently. It helps to deconstruct the behavior of functions resulting from multiplication, a common scenario in calculus.