Problem 55
Question
Find the intervals where \(f(x)=(\log x) / x\) is increasing and where it is decreasing.
Step-by-Step Solution
Verified Answer
The function \(f(x) = (\log(x) / x)\) is increasing on the interval (0, e) and decreasing on the interval (e, +∞). The function is not defined for x ≤ 0.
1Step 1: Calculate the derivative of f(x) using the quotient rule
The quotient rule states that for two functions u(x) and v(x), the derivative of their quotient (u/v) is given by: \((u/v)' = (vu' - uv') / v^2\).
For f(x) = log(x) / x, we have:
\(u(x) = \log(x)\), so \(u'(x) = 1/x\)
\(v(x) = x\), so \(v'(x) = 1\)
Now we can apply the quotient rule:
\(f'(x) = \frac{x(1/x) - \log(x)(1)}{x^2}\)
Simplify the expression:
\(f'(x) = \frac{1 - \log(x)}{x^2}\)
2Step 2: Identify critical points of f'(x)
To find the critical points, we look for where f'(x) = 0 or f'(x) is undefined.
Since f'(x) has a denominator, x = 0 is a critical point.
Now we solve for when f'(x) = 0:
\(\frac{1 - \log(x)}{x^2} = 0\)
\(1 - \log(x) = 0\)
\(\log(x) = 1\)
By the definition of logarithm, we get x = e.
So, we have two critical points, x = 0 and x = e.
3Step 3: Test intervals around critical points to determine increasing or decreasing
Now we test the sign of f'(x) in the intervals around the critical points. We will choose one test point between each pair of critical points.
1. Interval: x < 0
Choose a test point x = -1 (f'(x) is not defined for x ≤ 0 as log(x) is not defined for nonpositive x values, so this interval is not relevant for our analysis)
2. Interval: 0 < x < e
Choose a test point x = 1; then,
\(f'(1) = \frac{1 - \log(1)}{1^2} = 1 > 0\), which means f(x) is increasing.
3. Interval: x > e
Choose a test point x = e^2; then,
\(f'(e^2) = \frac{1 - \log(e^2)}{(e^2)^2} = \frac{1 - 2}{e^4} < 0\), which means f(x) is decreasing.
4Step 4: Write the final answer
Based on our analysis, f(x) = (\(\log(x) / x\)) is increasing on the interval (0, e) and decreasing on the interval (e, +∞). The function is not defined for x ≤ 0.
Key Concepts
Critical PointsInterval AnalysisIncreasing and Decreasing Functions
Critical Points
Critical points are special points on the graph of a function where its derivative is either zero or undefined. These points are important because they often relate to the peaks, troughs, or potentially flat regions of the graph. Understanding critical points helps us analyze the behavior of the function.
To find critical points for the function \( f(x) = \frac{\log(x)}{x} \), we need to find where the derivative \( f'(x) \) equals zero or is undefined. First, we derive the function using the quotient rule. The derivative \( f'(x) = \frac{1 - \log(x)}{x^2} \) equals zero at points where the numerator, \( 1 - \log(x) \), is zero. This results in \( \log(x) = 1 \), leading to \( x = e \) as a critical point.
Additionally, we must consider where the derivative is undefined. In this case, since the logarithm is undefined for non-positive \( x \), the critical point at \( x = 0 \) does not apply within the scope of positive real numbers.
To find critical points for the function \( f(x) = \frac{\log(x)}{x} \), we need to find where the derivative \( f'(x) \) equals zero or is undefined. First, we derive the function using the quotient rule. The derivative \( f'(x) = \frac{1 - \log(x)}{x^2} \) equals zero at points where the numerator, \( 1 - \log(x) \), is zero. This results in \( \log(x) = 1 \), leading to \( x = e \) as a critical point.
Additionally, we must consider where the derivative is undefined. In this case, since the logarithm is undefined for non-positive \( x \), the critical point at \( x = 0 \) does not apply within the scope of positive real numbers.
Interval Analysis
Interval analysis is used to determine the behavior of a function between critical points by evaluating the sign of its derivative.
This allows us to assess whether the function is increasing or decreasing within these intervals. After finding the critical points, our derivative \( f'(x) = \frac{1 - \log(x)}{x^2} \) leads us to explore two main intervals: \( (0, e) \) and \( (e, \infty) \). We exclude \( x \leq 0 \) as the logarithmic function is undefined for non-positive values.
To evaluate the sign of the derivative within the interval \( (0, e) \), we choose a test point \( x = 1 \). Plugging into the derivative \( f'(x) \), we find that it is positive, indicating the function is increasing in \( (0, e) \).
On testing \( x = e^2 \) for \( (e, \infty) \), \( f'(x) \) turns out negative, showing the function decreases on this interval.
This allows us to assess whether the function is increasing or decreasing within these intervals. After finding the critical points, our derivative \( f'(x) = \frac{1 - \log(x)}{x^2} \) leads us to explore two main intervals: \( (0, e) \) and \( (e, \infty) \). We exclude \( x \leq 0 \) as the logarithmic function is undefined for non-positive values.
To evaluate the sign of the derivative within the interval \( (0, e) \), we choose a test point \( x = 1 \). Plugging into the derivative \( f'(x) \), we find that it is positive, indicating the function is increasing in \( (0, e) \).
On testing \( x = e^2 \) for \( (e, \infty) \), \( f'(x) \) turns out negative, showing the function decreases on this interval.
Increasing and Decreasing Functions
An increasing function rises as you move along the x-axis from left to right, while a decreasing function falls during this progression. The derivative tells us the function's slope at any point.
If \( f'(x) > 0 \) for an interval, then the function is increasing there. Conversely, if \( f'(x) < 0 \) in the interval, the function is decreasing. Based on the interval analysis above:
If \( f'(x) > 0 \) for an interval, then the function is increasing there. Conversely, if \( f'(x) < 0 \) in the interval, the function is decreasing. Based on the interval analysis above:
- On the interval \( (0, e) \), since \( f'(x) > 0 \), the function \( f(x) = \frac{\log(x)}{x} \) is increasing.
- On the interval \( (e, \infty) \), since \( f'(x) < 0 \), the function is decreasing.
Other exercises in this chapter
Problem 55
In Exercises \(55-58\), plot the graph of the function. $$ f(t)=\frac{\sqrt{t^{2}+1}}{t-1} $$
View solution Problem 55
Find the horizontal and vertical asymptotes of the graph of the function. Do not sketch the graph. $$ f(t)=\frac{t^{2}-2}{t^{2}-4} $$
View solution Problem 55
In Exercises \(41-60\), find the absolute maximum and absolute minimum values, if any, of the function. $$ f(x)=2+3 \sin 2 x \text { on }\left[0, \frac{\pi}{2}\
View solution Problem 56
In Exercises \(55-58\), plot the graph of the function. $$ f(x)=\frac{x^{2}+x}{3 x^{2}+x-1} $$
View solution