Problem 30 - Calculus Early Transcendentals

Question

More graphing Make a complete graph of the following functions. If an interval is not specified, graph the function on its domain. Use a graphing utility to check your work. $$g(x)=x^{2} \ln x$$

Step-by-Step Solution

Answer

Question: Sketch the graph of the function $$g(x)=x^2 \ln x$$ and identify its behavior. Solution: The graph of the function $$g(x)=x^2 \ln x$$ has the following characteristics: 1. The domain is $$(0, +\infty)$$. 2. It has a critical point at $$x = e^{-\frac{1}{2}}$$. 3. The function is decreasing for $$0 < x < e^{-\frac{1}{2}}$$ and increasing for $$x > e^{-\frac{1}{2}}$$. 4. The graph starts from the positive y-axis (as x approaches 0), decreases till the critical point, and then starts increasing.

Step 1: Identify the domain of the function

The function $$g(x) = x^2 \ln x$$ is a product of a quadratic function $$x^2$$ and the natural logarithm function $$\ln x$$. The quadratic function has the domain of all real numbers, while the natural logarithm function has the domain $$(0, +\infty)$$. Therefore, the domain of $$g(x)$$ is $$(0, +\infty)$$.

Step 2: Find critical points

To find the critical points of the function, we need to find the derivative of $$g(x)$$ and set it equal to zero. The derivative of $$g(x)$$ is: $$g'(x) = \frac{d}{dx} (x^2 \ln x)$$ We will use the product rule: $$\frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v'$$ Here, $$u(x) = x^2$$ and $$v(x) = \ln x$$. After differentiating and substituting, we get: $$g'(x) = (2x) \cdot (\ln x) + (x^2) \cdot (\frac{1}{x})$$ Which simplifies to: $$g'(x) = 2x \ln x + x$$ Now, let's set the derivative equal to zero and solve for $$x$$: $$g'(x) = 0$$ $$2x \ln x + x = 0$$ Factor out $$x$$: $$x (2 \ln x + 1) = 0$$ This equation is true if either $$x = 0$$ or $$2 \ln x + 1 = 0$$. However, as x is strictly greater than 0 in the domain, x=0 is not a solution. The other possibility is: $$2 \ln x + 1 = 0$$ $$\ln x = -\frac{1}{2}$$ $$x = e^{-\frac{1}{2}}$$ So, the function has a critical point at $$x = e^{-\frac{1}{2}}$$.

Step 3: Analyze the function's behavior

We have identified that the function has a critical point at $$x = e^{-\frac{1}{2}}$$. Now, let's analyze the behavior of $$g(x)$$ in different regions of its domain using the sign of the derivative. 1. If $$0 < x < e^{-\frac{1}{2}}$$, then $$\ln x < -\frac{1}{2}$$ and $$2 \ln x + 1 < 0$$, meaning that $$g'(x) < 0$$. So, $$g(x)$$ is decreasing in this interval. 2. If $$x > e^{-\frac{1}{2}}$$, then $$\ln x > -\frac{1}{2}$$ and $$2 \ln x + 1 > 0$$, meaning that $$g'(x) > 0$$. So, $$g(x)$$ is increasing in this interval.

Step 4: Sketch the graph

Using the information from Steps 1-3, we can now sketch the graph of $$g(x) = x^2 \ln x$$ on its domain $$(0, + \infty)$$: 1. The function starts from the positive y-axis (as x approaches 0) and is decreasing till it reaches the critical point at $$x = e^{-\frac{1}{2}}$$. 2. After the critical point, the function starts increasing. This gives us a rough sketch of the function. Use a graphing utility for a more accurate representation.

Step 5: Verify the graph using a graphing utility

To check our work, we can use a graphing utility (such as Desmos, GeoGebra, or a graphing calculator) to graph $$g(x) = x^2 \ln x$$ and verify that the graph matches our explanation.

Key Concepts

Graphing Functions

Graphing functions helps us visualize mathematical relationships, allowing us to better understand how a function behaves across its domain. The function given, \(g(x) = x^2 \ln x\), combines a quadratic term \(x^2\) and the natural logarithm function \(\ln x\). Graphing it requires understanding both the individual components and how they interact over the function's domain.

Here's a step-by-step guide on how to effectively approach graphing such functions:

• **Identify the domain**: This determines the allowable inputs for the function, helping you set the range for the graph's horizontal axis.
• **Find critical points**: These points can indicate where the function changes direction—either from increasing to decreasing or vice versa.
• **Use derivative analysis**: The slope of the function at various points can tell you about its increasing or decreasing nature.
• **Sketch based on findings**: Begin sketching with an understanding of the function's start point, critical points, and general direction.

Visualization with a graphing tool can serve as a final confirmation to ensure the graph aligns with mathematical predictions.

Critical Points

Critical points help us understand where a function's trend changes from increasing to decreasing, or vice versa. For \(g(x) = x^2 \ln x\), these are found by calculating its derivative and solving for zero.

To find the critical points:

• **Compute the derivative**: Apply the product rule here since \(g(x)\) includes a product of functions.
• **Find where the derivative is zero or undefined**: This reveals potential critical points.
• **Solve for the unknown**: Here, this involved solving \(2x \ln x + x = 0\).
• **Evaluate within the domain**: Consider only those solutions that fall within the domain \((0, +\infty)\).

For \(g(x)\), the process revealed a critical point at \(x = e^{-\frac{1}{2}}\), which is essential for mapping its behavior along the graph.

Derivative

The derivative of a function gives us a powerful tool to determine its slope at any given point, helping us understand its rates of change. For \(g(x) = x^2 \ln x\), finding the derivative provides insight into how the function increases or decreases.

Here's the approach to find and manage derivatives in this context:

• **Use the product rule for differentiation**: Given that \(g(x)\) is a product of \(x^2\) and \(\ln x\), the product rule \, \( \frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v' \, \) is used.
• **Differentiate each component**: Identify \(u(x) = x^2\) and \(v(x) = \ln x\), then calculate \(u'\) and \(v'\).
• **Combine results**: Substitute back into the product rule formula, simplifying to obtain \(g'(x) = 2x \ln x + x\).

This process not only reveals where critical points lie but also helps to understand the general behavior of the function in its domain.

Domain of a Function

Understanding the domain of a function is crucial as it defines the set of inputs for which the function exists. The function \(g(x) = x^2 \ln x\) combines components with different natural domains:

• For \(x^2\), the domain is all real numbers (\(-\infty, +\infty\)). However, \(x^2\) is always non-negative for real inputs, simplifying visualization.
• For \(\ln x\), the domain is \((0, +\infty)\) because the natural logarithm function only accepts positive numbers.

To determine the domain of \(g(x)\), we take the intersection of these domains: \((0, +\infty)\). This reflects only the valid \(x\) values for which \(g(x)\) is defined.
An understanding of a function's domain helps to guide graphing efforts and ensures that any critical points or behaviors identified are relevant to the function's true landscape.