Problem 9 - Calculus with Concepts in Calculus

Question

Find the intervals on which the graph of the function is concave upward and those on which it is concave downward. $$ f(x)=x \ln x $$

Step-by-Step Solution

Answer

The function is concave upward on \((0, \infty)\). There are no intervals where it's concave downward.

Step 1: Find the First Derivative

To analyze the concavity of the function, we first need to find the first derivative of the given function \( f(x) = x \ln x \). Let's use the product rule here. The product rule states \((u v)' = u'v + uv'\). Here, let \( u = x \) and \( v = \ln x \). Then: \[ f'(x) = (x)' \ln x + x (\ln x)' \] Therefore, \[ f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \]

Step 2: Find the Second Derivative

Next, we differentiate \( f'(x) = \ln x + 1 \) to find the second derivative, which will allow us to analyze concavity. The derivative of \( \ln x \) is \( \frac{1}{x} \) and the derivative of a constant is 0. So, \[ f''(x) = \frac{1}{x} \]

Step 3: Determine Concavity from Second Derivative

The second derivative \( f''(x) = \frac{1}{x} \) indicates the nature of concavity. The function is concave upward when \( f''(x) > 0 \) and concave downward when \( f''(x) < 0 \). - For \( \frac{1}{x} > 0 \), \( x > 0 \). Therefore, \( f(x) \) is concave upward on the interval \((0, \infty)\). - Since \( \frac{1}{x} \) does not become negative for any value of \( x \) in the defined domain, there are no intervals where \( f(x) \) is concave downward.

Key Concepts

First Derivative

Understanding the concept of the first derivative is crucial when analyzing the properties of a function, such as finding where it's increasing or decreasing. In this particular exercise, we're using it to find where the graph changes its slope. For the function given, \(f(x) = x \ln x\), we apply a mathematical technique known as the product rule. The product rule helps us take the derivative of a product of two functions. It states that if you have a function \( f(x) = u(x) \cdot v(x) \), then the derivative \( f'(x) \) is \( u'(x) v(x) + u(x) v'(x) \). In our case, \( u = x \) and \( v = \ln x \). Here’s how it looks when broken down:

• Calculate the derivative of \(x\), which is 1.
• Calculate the derivative of \(\ln x\), which is \(\frac{1}{x}\).

Using these, we apply the product rule to get \( f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \), which gives us the slope of the original function at any point \(x\). This first derivative indicates where the function is leveling out or changing more rapidly.

Second Derivative

Once we have the first derivative, we move on to find the second derivative. This derivative provides information about the concavity of the function, telling us how the slope itself is changing.For the first derivative found earlier \( f'(x) = \ln x + 1 \), we need to compute its derivative. The process is straightforward:

• The derivative of \(\ln x\) is \(\frac{1}{x}\).
• The derivative of the constant 1 is 0.

Thus, we find that \( f''(x) = \frac{1}{x} \), which is crucial. The sign of this second derivative determines the concavity. Specifically, if \( f''(x) > 0 \), the function is concave upward, and if \( f''(x) < 0 \), it is concave downward.Here, since \( \frac{1}{x} > 0 \) for any \(x > 0\), we conclude that the function is concave upward on the interval \((0, \infty)\).

Product Rule

The product rule can initially appear daunting, but it's a powerful tool for differentiation, especially when dealing with products of functions as in this exercise.When you have two separate functions multiplied together, like \(x\) and \(\ln x\) in the function \(f(x) = x \ln x\), their product's derivative can be found using the product rule: \[(uv)' = u'v + uv'\] For \(u = x\) and \(v = \ln x\), we go step by step:

• First, derive \(u\) to get \(1\), and \(v\) to get \(\frac{1}{x}\).
• Then apply these derivatives in the rule: \( u'v + uv' = 1\cdot\ln x + x\cdot\frac{1}{x} \).

Carrying these calculations leads us to \(f'(x) = \ln x + 1\). Understanding and applying the product rule means you can easily tackle more complex differentiation problems involving multiple functions.