Problem 32 - Calculus

Question

Estimate the instantaneous rate of change of the function \(f(x)=x \ln x\) at \(x=1\) and at \(x=2 .\) What do these values suggest about the concavity of the graph between 1 and \(2 ?\)

Step-by-Step Solution

Answer

The function has positive slopes at both \(x=1\) and \(x=2\), suggesting concavity up.

Step 1: Differentiation of the Function

First, find the derivative of the function \(f(x) = x \ln x\). We will use the product rule here, which states \((uv)' = u'v + uv'\), where \(u = x\) and \(v = \ln x\). Therefore, we have:\[ f'(x) = \frac{d}{dx}(x) \cdot \ln x + x \cdot \frac{d}{dx}(\ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1.\]

Step 2: Evaluate the Derivative at x=1

Substitute \(x = 1\) into the derivative \(f'(x) = \ln x + 1\):\[ f'(1) = \ln 1 + 1 = 0 + 1 = 1.\]So, the instantaneous rate of change of the function at \(x = 1\) is 1.

Step 3: Evaluate the Derivative at x=2

Substitute \(x = 2\) into the derivative \(f'(x) = \ln x + 1\):\[ f'(2) = \ln 2 + 1.\]Using the approximate value \(\ln 2 \approx 0.693\), we get:\[ f'(2) = 0.693 + 1 = 1.693.\]So, the instantaneous rate of change of the function at \(x = 2\) is approximately 1.693.

Step 4: Analyze the Concavity

Since \(f'(x)\) is increasing from 1 to 2 (1 to 1.693), the function \(f(x)\) is concave up between the interval \(x = 1\) and \(x = 2\). This suggests that the graph is bending upwards in this interval.

Key Concepts

Differentiation

Differentiation is like the magic wand of calculus that lets us understand how a function changes at any point. It's all about finding the derivative, which gives us the rate of change of a function at any particular instant. Imagine you're driving a car—if you want to know how fast you're going right at a specific moment, you're essentially looking for the car's instantaneous speed. Similarly, differentiation helps us find out how a function is behaving at a particular point.

For the function given in the exercise, which is \(f(x) = x \ln x\), differentiation lets us figure out how \(f(x)\) is changing as \(x\) varies. To differentiate this function, we use something known as the "Product Rule" which is super handy when you have two functions being multiplied together, like \(x\) and \(\ln x\) here. Understanding differentiation paves the way for deeper insights into the behavior of functions. This helps not just in mathematics but also in physics, engineering, and beyond.

Product Rule

The Product Rule is a fantastic tool in calculus used when differentiating products of two functions. In simple terms, if you have a function that is the product of two smaller functions, this rule tells you how to find its derivative. Let’s break it down a bit more.

We use the Product Rule when we have a function like \(uv\), where \(u\) and \(v\) are both functions of \(x\). The rule states that the derivative of \(uv\) is \((uv)' = u'v + uv'\). What this means is, you take the derivative of the first function \(u\) and multiply it by the second function \(v\), then add the first function \(u\) multiplied by the derivative of the second function \(v'\).

• Imagine you have the function \(f(x) = x \ln x\)
• Here, \(u = x\) (so \(u' = 1\)) and \(v = \ln x\) (so \(v' = \frac{1}{x}\))
• Applying the Product Rule gives us \(f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\)

It's like having a recipe for finding the change at every point of the multiplied function.

Concavity

Concavity tells us how a curve bends. Does it smile or frown? When you look at the graph of a function, concavity helps you understand whether the graph opens upwards or downwards.

For our specific function \(f(x) = x \ln x\), after finding the derivative \(f'(x)\), we analyzed it for different values of \(x\) (like \(x = 1\) and \(x = 2\)). If the derivative increases as \(x\) increases, it suggests the graph is bending upwards, meaning it's concave up. Just imagine a bowl facing up—if you pour water in it, the water stays inside.

In this exercise, between \(x = 1\) and \(x = 2\), we noticed that the value of \(f'(x)\) went from 1 to approximately 1.693. Since the derivative is increasing, it indicates that the function is experiencing concave upward behavior in this interval. Understanding concavity is key because it doesn’t just tell you the direction in which the graph curves, but also provides insights into things like maximum and minimum points.