Problem 68 - Algebra and Trigonometry

Question

\(65-70\) Draw the graph of the function in a suitable viewing rectangle and use it to find the domain, the asymptotes, and the local maximum and minimum values. $$ y=x(\ln x)^{2} $$

Step-by-Step Solution

Answer

Domain: \((0, \infty)\). Asymptotes: None. Local Max: \( y = \frac{4}{e^2} \) at \( x = \frac{1}{e^2} \). Local Min: \( y = 0 \) at \( x = 1 \).

Step 1: Understanding the Function

The given function is \( y = x(\ln x)^2 \). This is a logarithmic function modified by \( x \), with notable changes occurring at values near the natural base and where \( x \) is positive.

Step 2: Finding the Domain

The domain is determined by where the logarithmic function is defined. Since \( \ln x \) is only defined for \( x > 0 \), the domain of the function is \((0, \infty)\).

Step 3: Identifying Asymptotes

The function does not have any vertical asymptotes, as \( \ln x \) is undefined only at \( x = 0 \), where the function is also undefined. There is no horizontal asymptote because the function tends to infinity as \( x \) approaches infinity.

Step 4: Finding Critical Points

To find critical points, differentiate the function. Let \( f(x) = x(\ln x)^2 \). The derivative \( f'(x) \) can be calculated using product and chain rules: \( \frac{d}{dx}(x) = 1 \) and the derivative \( \frac{d}{dx}(\ln x)^2 = 2\ln x \cdot \frac{1}{x} \). Therefore, \( f'(x) = (\ln x)^2 + 2\ln x \). Solve \( f'(x) = 0 \) to get critical points.

Step 5: Solving for Critical Points

Set \( (\ln x)^2 + 2\ln x = 0 \). Let \( u = \ln x \). Then \( u^2 + 2u = 0 \), which factors to \( u(u + 2) = 0 \), giving \( u = 0 \) or \( u = -2 \). Reverse substitution gives \( \ln x = 0 \) or \( \ln x = -2 \). Solving for \( x \), we get \( x = e^0 = 1 \) and \( x = e^{-2} = \frac{1}{e^2} \).

Step 6: Analyzing the Critical Points

Evaluate the function at \( x = 1 \) and \( x = \frac{1}{e^2} \) to find local maxima/minima. At \( x = 1 \), \( f(1) = 1 \cdot (\ln(1))^2 = 0 \), a minimum. At \( x = \frac{1}{e^2} \), find \( f(\frac{1}{e^2}) = \frac{1}{e^2}(-2)^2 = \frac{4}{e^2} \), so this is a local maximum due to the zero derivative and derivative sign change.

Step 7: Graphing the Function

Graph \( y = x(\ln x)^2 \) in a suitable viewing window such as from \( x = 0.1 \) to \( x = 3 \), where changes are easily observed. The function should rise then fall to a minimum at \( x = 1 \), then rise again.

Key Concepts

Function Domain

Understanding the domain of a function is essential because it tells us where the function is defined and can be evaluated. For the given function, \( y = x(\ln x)^2 \), the presence of \( \ln x \) is crucial to determining the domain.- **Logarithmic Domain:** The natural logarithm \( \ln x \) is only defined for values of \( x \) that are greater than zero. This is because the logarithm of a non-positive number is undefined in the real number system. Therefore, any expression involving \( \ln x \) will similarly have its domain restricted to positive values of \( x \).- **Determined Domain:** Consequently, the domain of the function \( y = x(\ln x)^2 \) is from 0 to infinity, expressed in interval notation as \((0, \infty)\).Remember that understanding the domain helps in analyzing and graphing the function properly. It ensures that you consider only values where the function actually exists.

Asymptotes

Asymptotes of a function indicate lines that the graph of the function approaches but never touches. They are important in analyzing the behavior of functions as the input values become very large or very small.- **Vertical Asymptotes:** These occur where the function is undefined and the function values grow very large in magnitude. For \( y = x(\ln x)^2 \), since \( \ln x \) is undefined at \( x = 0 \), you might suspect a vertical asymptote there. However, the function itself also does not exist for \( x \leq 0 \), which means no vertical asymptote actually forms.- **Horizontal Asymptotes:** These typically indicate a leveling off of the function as \( x \) becomes very large. For this function, as \( x \) approaches infinity, \( x(\ln x)^2 \) also tends toward infinity. Therefore, there is no horizontal asymptote.By recognizing the absence of asymptotes, we see that the behavior near zero and infinity must be expressed in terms of function growth rather than approach to a line.

Critical Points

Critical points help identify where a function's slope changes, indicating potential maxima, minima, or points of inflection.- **Finding Critical Points:** To find them for \( y = x(\ln x)^2 \), we differentiate the function. Using the product and chain rules, the derivative \( f'(x) = (\ln x)^2 + 2\ln x \) is derived. A critical point occurs where \( f'(x) = 0 \).- **Solving Derivative Equation:** Let \( u = \ln x \), and \( u^2 + 2u = 0 \) simplifies to \( u(u+2) = 0 \). Solving this, \( u = 0 \) or \( u = -2 \). Reverse substitution gives \( \ln x = 0 \) (thus \( x = 1 \)) and \( \ln x = -2 \) (thus \( x = \frac{1}{e^2} \)).By setting the derivative to zero and solving, we can successfully find critical points that will help us in assessing intervals of increase and decrease.

Local Extrema

Local extrema refer to the highest or lowest points in a region of a function's graph. Analyzing these points can help describe the function's behavior.- **Identifying Local Extrema:** Once critical points have been found, you can evaluate the function at these points to decide if they are a maximum, minimum, or neither. At \( x = 1 \), \( f(1) = 1 \cdot (\ln 1)^2 = 0 \); this point is a local minimum because the value is lower compared to values around it. At \( x = \frac{1}{e^2} \), the function attains a local maximum as \( f\left(\frac{1}{e^2}\right) = \frac{4}{e^2} \).- **Analyzing Behavior:** By examining the sign of \( f'(x) \) around these points, you can confirm their status as maxima or minima. The slope changes from negative to positive at \( x = 1 \), indicating a minimum.Identifying local extrema offers insights into the shape and symmetry of the graph, showing where the function has peaks and valleys.