Chapter 6: Problem 48
Behavior at the origin Using calculus and accurate sketches, explain how the graphs of \(f(x)=x^{p} \ln x\) differ as \(x \rightarrow 0\) for \(p=\frac{1}{2}, 1,\) and 2
Short Answer
Expert verified
For \(p = \frac{1}{2}\), the function approaches 0 as \(x \rightarrow 0^+\), and its graph is increasing and steep near the origin.
For \(p = 1\), the function approaches 0 as \(x \rightarrow 0^+\), and its graph is increasing and less steep near the origin compared to the case for \(p = \frac{1}{2}\).
For \(p = 2\), the function approaches 0 as \(x \rightarrow 0^+\), and its graph is increasing and less steep near the origin compared to the case for \(p = 1\).
Step by step solution
01
Analyze the function for \(p = \frac{1}{2}\)
To analyze the function \(f(x) = x^{1/2} \ln{x}\) as \(x \rightarrow 0\), we need to find the limit as x approaches 0. Since \(\ln{0}\) is undefined, we will use L'Hôpital's rule to evaluate the limit:
$$\lim_{x\to 0^+} \frac{\ln{x}}{x^{-1/2}}$$
Applying L'Hôpital's rule, we need to find the derivatives of the numerator and denominator with respect to \(x\):
$$\lim_{x\to 0^+} \frac{\frac{1}{x}}{-\frac{1}{2}x^{-3/2}} = \lim_{x\to 0^+} \frac{-2x^{1/2}}{1}$$
Since the limit as \(x \rightarrow 0^+\) is 0, the function approaches 0 as x approaches 0.
To get an idea of the shape of the graph, we can find its derivative:
$$f'(x) = \frac{d}{dx}(x^{1/2} \ln{x}) = x^{1/2} \frac{1}{x} + \frac{1}{2}x^{-1/2} \ln{x} = \frac{1}{\sqrt{x}} + \frac{\ln{x}}{2\sqrt{x}}$$
As \(x \rightarrow 0^+\), the first term approaches infinity while the second term approaches negative infinity. Therefore the graph will be increasing and steep near the origin.
02
Analyze the function for \(p = 1\)
Now let's analyze the function \(f(x) = x \ln{x}\) as \(x \rightarrow 0\). The limit to find here is:
$$\lim_{x\to 0^+} x \ln{x}$$
This limit evaluates to 0 directly, so the function also approaches 0 as \(x\) approaches 0.
For the shape of the graph, we will find its derivative:
$$f'(x) = \frac{d}{dx}(x \ln{x}) = \ln{x} + 1$$
As \(x \rightarrow 0^+\), the derivative approaches negative infinity. Therefore the graph will be increasing and less steep near the origin compared to the case for \(p = \frac{1}{2}\).
03
Analyze the function for \(p = 2\)
Finally, let's analyze the function \(f(x) = x^2 \ln{x}\) as \(x \rightarrow 0\). The limit to find is:
$$\lim_{x\to 0^+} x^2 \ln{x}$$
This limit evaluates to 0 directly, so the function also approaches 0 as \(x\) approaches 0.
For the shape of the graph, we will find its derivative:
$$f'(x) = \frac{d}{dx}(x^2 \ln{x}) = 2x \ln{x} + x^2 \frac{1}{x} = 2x \ln{x} + x$$
As \(x \rightarrow 0^+\), the derivative approaches 0. This means that the graph will be increasing and less steep near the origin than the case for \(p=1\).
04
Conclusion
Now we know the behavior of the function \(f(x) = x^p \ln{x}\) as \(x \rightarrow 0\) for each value of \(p\):
- For \(p = \frac{1}{2}\), the function approaches 0 as \(x \rightarrow 0^+\), and its graph is increasing and steep near the origin.
- For \(p = 1\), the function approaches 0 as \(x \rightarrow 0^+\), and its graph is increasing and less steep near the origin compared to the case for \(p = \frac{1}{2}\).
- For \(p = 2\), the function approaches 0 as \(x \rightarrow 0^+\), and its graph is increasing and less steep near the origin compared to the case for \(p = 1\).
Using this information, one can accurately sketch the graphs to visualize their differences as \(x \rightarrow 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hôpital's rule
L'Hôpital's Rule is a powerful tool in calculus used to calculate limits that are difficult to address directly. This rule helps us find limits for functions that approach an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) as \( x \to a \). It's a way to transform complicated limits into something easier to manage.
To apply L'Hôpital's Rule, follow these steps:
To apply L'Hôpital's Rule, follow these steps:
- Verify that your limit is of an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
- Differentiate the numerator and the denominator separately.
- Re-evaluate the limit.
Behavior at the origin
Understanding the behavior of functions as \( x \to 0 \) is crucial in many calculus problems. It helps us sketch curves and predict how functions will behave near the origin. When analyzing a function's behavior at the origin, consider:
The steepness of the graph at the origin differed based on \( p \):
- Limits: Ensure that the function is approaching a defined value as \( x \to 0 \). In our exercise, regardless of \( p \), the limit evaluated to 0.
- Steepness: Analyze how quickly or slowly the function increases or decreases. This is often done by finding the function's derivative.
The steepness of the graph at the origin differed based on \( p \):
- \( p = \frac{1}{2} \): The graph is very steep, shooting upwards quickly.
- \( p = 1 \): The steepness is less pronounced compared to \( p = \frac{1}{2} \).
- \( p = 2 \): The graph rises even more gently near the origin.
Function analysis
Function analysis involves breaking down a function into understandable parts to understand its characteristics and predict its behavior. This includes:
Next, by evaluating the derivatives, information about the function's steepness was extracted, vital for sketching graphs and understanding how the behavior changes with \( p \). This analysis allows us to:
- Identifying its limit behavior at crucial points like \( x \to 0 \).
- Evaluating its first derivative to understand the rate of change and determine steepness.
- Using derivative information to sketch graphs and infer concavity or points of inflection.
Next, by evaluating the derivatives, information about the function's steepness was extracted, vital for sketching graphs and understanding how the behavior changes with \( p \). This analysis allows us to:
- Spot critical points.
- Understand increasing or decreasing behavior.
- Analyze potential asymptotes or boundaries.