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Magazine Chapter 4: Problem 32
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. $$f(x)=(\ln x) / x^{2}$$
Short Answer
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Question: Sketch a graph of the function \(f(x) = \frac{\ln x}{x^2}\), including its domain, asymptotes, increasing and decreasing intervals, and concavity.
Step by step solution
01
Determine the Domain of the Function
First, let's find the domain of the function \(f(x) = \frac{\ln x}{x^2}\). The domain consists of all x values for which the function is defined. Since we cannot have the natural logarithm of a non-positive x value, the domain of the function is \((0,\infty)\).
02
Find the Critical Points
Next, find the x-intercepts and y-intercepts. The x-intercept occurs when \(f(x) = 0\), and the y-intercept occurs when \(x = 0\).
2.1. x-intercepts: Solve \(f(x) = 0\) for x. However, in the case of our function, this equation cannot be solved analytically, so x-intercepts don't play a significant role in creating the graph.
2.2. y-intercept: Since the domain of our function begins from \(x>0\), there is no y-intercept point in this function.
03
Find the Asymptotes
3.1. Vertical asymptotes: Vertical asymptotes occur when the denominator is equal to zero or the function approaches infinity. In this case, the denominator is never zero; however, the natural logarithm approaches negative infinity as x approaches zero, so there's a vertical asymptote at x=0.
3.2. Horizontal asymptotes: Horizontal asymptotes occur when the function approaches a certain value as x approaches positive or negative infinity. As x approaches infinity, the natural logarithm grows much slower than \(x^2\), so \(f(x)\) approaches 0. Thus, there is a horizontal asymptote at y=0.
04
Determine the Increasing and Decreasing Intervals
To find where the function is increasing or decreasing, we need to look at the first derivative of the function:
\(f'(x) = \frac{d}{dx}(\frac{\ln x}{x^2}) = \frac{1 - 2\ln x}{x^3}\)
The critical points occur when \(f'(x) = 0\) or when the derivative doesn't exist:
\(f'(x) = 0\) when \(1 - 2\ln x = 0\) or \(x = e^{1/2}\)
The function is increasing on the interval \((0, e^{1/2})\) and decreasing on the interval \((e^{1/2}, \infty)\).
05
Analyze the Concavity
Now, we determine the concavity by looking at the second derivative:
\(f''(x) = \frac{d^2}{dx^2}(\frac{\ln x}{x^2}) = \frac{6 - 6\ln x}{x^4}\)
The critical points occur when \(f''(x) = 0\) or when the second derivative doesn't exist:
\(f''(x) = 0\) when \(6 - 6\ln x = 0\) or \(x = e^1\)
The function is concave up on the interval \((0, e^1)\) and concave down on the interval \((e^1, \infty)\).
Now, with this information, we can sketch a graph of the function \(f(x) = \frac{\ln x}{x^2}\). The graph should display the domain, asymptotes, increasing and decreasing intervals, and concavity. Additionally, you may use a graphing utility to verify your sketch.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of a Function
The domain of a function defines the set of all possible inputs for that function. In simpler terms, it tells us the "legal" x-values that we can use while keeping the function defined.
For our function, \(f(x)=\frac{\ln x}{x^2}\), the domain is determined by understanding the restrictions of the natural logarithm and the division rules.
For our function, \(f(x)=\frac{\ln x}{x^2}\), the domain is determined by understanding the restrictions of the natural logarithm and the division rules.
- The natural logarithm, \(\ln x\), requires that x be greater than 0 because the logarithm of a zero or negative number is not defined. This restricts our domain to \((0, \infty)\).
- The denominator cannot be zero as you cannot divide by zero. However, since \(x^2 > 0\) for all x in our domain, this is not a concern here.
Asymptotes
Asymptotes are lines that the graph of a function approaches but never quite touches.
There are two main types relevant to this function: vertical and horizontal.
There are two main types relevant to this function: vertical and horizontal.
- Vertical Asymptotes: - This occurs where the function tends towards infinity. Since \(\ln x\) approaches negative infinity as \(x\) approaches 0 from the right, there's a vertical asymptote at \(x=0\).
- Horizontal Asymptotes: - These occur when the function approaches a limit as \(x\) approaches infinity. - For our function, as \(x\) becomes infinitely large, the \(\ln x\) term grows much slower than the \(x^2\) term, making \(f(x)\) approach 0. Hence, there's a horizontal asymptote at \(y=0\).
Critical Points
Critical points are where the function's slope is zero or undefined, indicating a potential maximum, minimum, or point of inflection.
To find these, we use the first derivative, \(f'(x)\).
To find these, we use the first derivative, \(f'(x)\).
- The calculation gives \(f'(x) = \frac{1 - 2\ln x}{x^3}\).
- A critical point happens where \(f'(x) = 0\), which simplifies to \(x = e^{1/2}\).
- These points help us understand where the function transitions from increasing to decreasing or vice versa.
- The function increases on \((0, e^{1/2})\) and decreases on \((e^{1/2}, \infty)\), making \(x = e^{1/2}\) a peak turning point.
Concavity
Concavity describes how the curve bends. A function can be either concave up (like a cup) or concave down (like a frown).
To determine the concavity of \(f(x)\), we examine the second derivative, \(f''(x)\).
To determine the concavity of \(f(x)\), we examine the second derivative, \(f''(x)\).
- The calculation yields \(f''(x) = \frac{6 - 6\ln x}{x^4}\).
- A point of inflection, where concavity changes, occurs when \(f''(x) = 0\). This simplifies to \(x = e^1\).
- Thus, the function is concave up on \((0, e^1)\) and concave down on \((e^1, \infty)\).
