Problem 25 - Calculus Single Variable

Question

Use the first derivative to determine the intervals on which the given function \(f\) is increasing and on which \(f\) is decreasing. At each point \(c\) with \(f^{\prime}(c)=0,\) use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=x \log _{2}(x) $$

Step-by-Step Solution

Answer

The function is decreasing on \((0, \frac{1}{e})\) and increasing on \((\frac{1}{e}, \infty)\). It has a local minimum at \(x = \frac{1}{e}\).

Step 1: Find the Derivative

First, we need to find the derivative of the function \(f(x)=x \log_2(x)\). We can rewrite the logarithm in a more convenient form for differentiation: \(f(x) = \frac{x \ln(x)}{\ln(2)}\). Now use the product rule, which is \((uv)' = u'v + uv'\), where \(u = x\) and \(v = \frac{\ln(x)}{\ln(2)}\). The derivative is:\[f'(x) = 1 \cdot \frac{\ln(x)}{\ln(2)} + x \cdot \frac{1}{x\ln(2)} = \frac{\ln(x)+1}{\ln(2)}.\]

Step 2: Solve for Critical Points

Set the derivative \(f'(x) = \frac{\ln(x)+1}{\ln(2)}\) to zero to find critical points:\[\ln(x) + 1 = 0.\]Solving for \(x\), we get \(\ln(x) = -1\), giving \(x = e^{-1} = \frac{1}{e}\). This is our critical point.

Step 3: Identify Intervals of Increase and Decrease

To determine where the function is increasing or decreasing, analyze the sign of \(f'(x)\) around the critical point \(x = \frac{1}{e}\). Consider intervals \((0, \frac{1}{e})\) and \((\frac{1}{e}, \infty)\).- For \(x < \frac{1}{e}\), choose \(x = \frac{1}{4}\): - \(\ln(\frac{1}{4}) + 1 < 0\) because \(\ln(\frac{1}{4}) < -1\). - So \(f'(x) < 0\), indicating \(f\) is decreasing.- For \(x > \frac{1}{e}\), choose \(x = 1\): - \(\ln(1) + 1 > 0\). - So \(f'(x) > 0\), indicating \(f\) is increasing.

Step 4: Apply the First Derivative Test

Using the First Derivative Test at \(x = \frac{1}{e}\):- \(f'(x)\) changes from negative to positive (decreasing to increasing).- Therefore, \(f(x)\) has a local minimum at \(x = \frac{1}{e}\).

Key Concepts

Critical Points

When analyzing functions, critical points play a key role in understanding their behavior. A critical point occurs wherever the derivative of a function is equal to zero or undefined. These points are essential as they often indicate where a function may have local minimums or maximums.

In terms of our example, we identified the function as \(f(x) = x \log_2(x)\). By differentiating using the product rule, we found the derivative to be \(f'(x) = \frac{\ln(x) + 1}{\ln(2)}\). Setting this equal to zero provides location coordinates for critical points. Solving \(\ln(x) + 1 = 0\), we determined that \(x = \frac{1}{e}\) is a critical point of the function.

Understanding the role of critical points allows us to use tests like the First Derivative Test to explore further characteristics of the function’s graph, such as identifying local extrema (i.e., local minimums and maximums). These are important for visualizing and comprehending the overall behavior and trends of the function.

Intervals of Increase and Decrease

Determining where a function increases or decreases can be achieved by analyzing the derivative's sign around critical points. This is crucial for understanding the function's overall shape and behavior. Once we have the derivative, assessing its sign shows us these intervals.

For our function \(f(x) = x \log_2(x)\), we found \(f'(x) = \frac{\ln(x) + 1}{\ln(2)}\). Using our previously identified critical point \(x = \frac{1}{e}\), we must determine how \(f'(x)\) behaves in the intervals surrounding this point.

• For \(x < \frac{1}{e}\), choosing a test point like \(x = \frac{1}{4}\): Here, \(\ln(\frac{1}{4}) + 1\) is negative, implying \(f'(x) < 0\). The function decreases in this interval.
• For \(x > \frac{1}{e}\), choosing a test point like \(x = 1\): Now, \(\ln(1) + 1\) is positive, implying \(f'(x) > 0\). Hence, the function increases here.

These intervals tell us exactly where the function exhibits increasing and decreasing trends.

Product Rule

To effectively differentiate certain functions, understanding and applying the Product Rule is necessary. The Product Rule is a fundamental rule in calculus used when differentiating products of two functions. If we have two functions, \(u(x)\) and \(v(x)\), their derivative is expressed as \((uv)' = u'v + uv'\). This method ensures that each component of the product is differentiated appropriately.

In our example, with \(f(x) = x \log_2(x)\), we adopted the Product Rule by setting \(u = x\) and \(v = \frac{\ln(x)}{\ln(2)}\). By applying the rule, we differentiated to find:

• \(u' = 1\)
• \(v' = \frac{1}{x\ln(2)}\)

Thus, the derivative \(f'(x) = 1 \cdot \frac{\ln(x)}{\ln(2)} + x \cdot \frac{1}{x\ln(2)}\) simplifies to \(\frac{\ln(x) + 1}{\ln(2)}\).
Understanding and mastering the Product Rule allows you to tackle more complex differentiation problems with confidence, enabling you to decompose and manage functions multiplied together with ease.