Problem 44 - Single Variable Calculus

Question

Find the relative extrema, if any, of the function. Use the Second Derivative Test, if applicable. $$ f(x)=x(\ln x)^{2} $$

Step-by-Step Solution

Answer

The function \(f(x) = x(\ln x)^2\) has a relative minimum at x = 1 and a relative maximum at \(x = e^{-2}\).

Step 1: Find the first derivative of the function

To find the first derivative of the function \(f(x) = x(\ln x)^{2}\), we will need to use the product rule. The product rule states that \(\frac{d}{dx}(uv) = u'v + uv'\), where u and v are functions of x. Let \(u = x\) and \(v = (\ln x)^{2}\), then First, differentiate u with respect to x: $$ u' = \frac{d}{dx}(x)= 1 $$ Then, differentiate v with respect to x via the chain rule: $$ v' = \frac{d}{dx}((\ln x)^{2}) = 2(\ln x)(\frac{1}{x}) = 2\frac{\ln x}{x} $$ Now, apply the product rule: $$ f'(x) = u'v + uv' = 1(\ln x)^{2} + x(2\frac{\ln x}{x}) = (\ln x)^{2} + 2\ln x $$

Step 2: Find the critical points

To find the critical points, we need to set the first derivative equal to zero and solve for x: $$ f'(x) = (\ln x)^{2} + 2\ln x=0 $$ Factor out the common term: $$ \ln x((\ln x)+2)=0 $$ The critical points are: $$ x = e^{0}\ \text{and}\ x=e^{-2} $$ So the critical points are x = 1 and x = e^{-2}.

Step 3: Find the second derivative

To apply the second derivative test, we need to find the second derivative of the function. Differentiate the first derivative with respect to x: $$ f''(x) = \frac{d}{dx}\left((\ln x)^2 + 2\ln x\right) $$ Apply the chain rule and sum rule of differentiation: $$ f''(x) = 2(\ln x)\left(\frac{1}{x}\right) + 2\left(\frac{1}{x}\right) = 2\frac{\ln x + 1}{x} $$

Step 4: Apply the Second Derivative Test

To apply the Second Derivative Test, we need to evaluate the second derivative at the critical points: x = 1 and x = e^{-2}. At x = 1: $$ f''(1) = 2\frac{\ln 1 + 1}{1} = 2(0+1)= 2 $$ Since \(f''(1) > 0\), the function has a relative minimum at x = 1. At x = e^{-2}: $$ f''(e^{-2}) = 2\frac{\ln e^{-2} + 1}{e^{-2}} = 2\frac{-2 + 1}{e^{-2}} = -\frac{2}{e^{-2}} $$ Since \(f''(e^{-2}) < 0\), the function has a relative maximum at x = e^{-2}.

Step 5: Final Answer

The relative extrema of the function \(f(x) = x(\ln x)^2\) are: Relative minimum at x = 1 and relative maximum at \(x = e^{-2}\).

Key Concepts

Relative Extrema

Relative extrema refer to the highest or lowest points on a function within a specific interval. They can be either a relative maximum or minimum. A point is a relative maximum if it is higher than the points immediately around it. Conversely, a relative minimum is lower than the points surrounding it. Understanding where a function reaches these relative highs and lows is crucial in calculus as it helps us understand the behavior and shape of the graph of the function.

To find these points, we often use tools like derivative tests, such as the First Derivative Test or the Second Derivative Test, which can provide information on the increasing or decreasing behavior of a function near critical points. In the given exercise, after computing the second derivative for the critical points, we identified that there is a relative minimum at \(x = 1\) and a relative maximum at \(x = e^{-2}\). By using these tests, students are able to not only locate these extrema but also understand their significance in the context of a mathematical model.

First Derivative

The first derivative of a function provides crucial insight into the function's characteristics, specifically its slope and direction. In simple terms, the first derivative, denoted as \(f'(x)\), tells us how the function \(f(x)\) is changing at any given point. Is it increasing, staying constant, or decreasing? This is fundamentally important when we want to analyze a function for relative extrema.

In the exercise provided, we found the first derivative \(f'(x) = (\ln x)^2 + 2\ln x\) using the product rule. By setting this derivative to zero, we initiated the process of finding critical points. At these points, the slope of the tangent to the curve is zero, indicating potential locations for relative maxima or minima. Understanding how to differentiate and set the first derivative to zero is the first major step in locating extrema.

Critical Points

Critical points are specific values of \(x\) where the first derivative of a function is zero or does not exist. These points are important because they are potential locations for relative extrema—either minima, maxima, or saddle points.

Finding critical points involves setting the first derivative \(f'(x)\) equal to zero and solving for \(x\). In the given example, we solved \((\ln x)^2 + 2\ln x = 0\) to find the critical points, where we identified \(x = 1\) and \(x = e^{-2}\). These critical points are key candidates we then examine further using the second derivative test to determine the nature of the extrema at these points. This approach helps us pinpoint specific areas to look for changes in behavior of the function. Being able to find and interpret critical points is an essential skill when dealing with calculus problems related to optimization and analysis.

Product Rule

The product rule is a fundamental technique in calculus for differentiating functions that are products of two simpler functions. When you have a function \(f(x) = u(x) \, v(x)\), where both \(u\) and \(v\) are functions of \(x\), you can't just differentiate \(f(x)\) by differentiating \(u\) and \(v\) separately and multiply.

Instead, the product rule states: \(\frac{d}{dx}(uv) = u'v + uv'\). This means you take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.

In the exercise, we applied the product rule to the function \(f(x) = x(\ln x)^2\). By setting \(u = x\) and \(v = (\ln x)^2\), we calculated the derivatives individually and then applied the product rule. This resulted in the first derivative \(f'(x) = (\ln x)^2 + 2\ln x\), which was then used to find the critical points of the function. Mastery of the product rule is essential for tackling complex calculus problems that involve products of functions.