Problem 28 - Calculus Single Variable

Question

Use the first derivative to determine the intervals on which the given function \(f\) is increasing and on which \(f\) is decreasing. At each point \(c\) with \(f^{\prime}(c)=0,\) use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=\ln (x) / x $$

Step-by-Step Solution

Answer

The function is increasing on \((0, e)\) and decreasing on \((e, \infty)\). Local maximum at \(x = e\).

Step 1: Find the first derivative of the function

To find the critical points of the function, we need to compute the first derivative of \( f(x) = \ln(x)/x \) using the quotient rule. The quotient rule is \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). Let \( u = \ln(x) \) and \( v = x \), then \( u' = \frac{1}{x} \) and \( v' = 1 \). Therefore, the first derivative is:\[ f'(x) = \frac{\frac{1}{x} \cdot x - \ln(x) \cdot 1}{x^2} = \frac{1 - \ln(x)}{x^2}. \]

Step 2: Identify critical points where f'(x) = 0

Set the first derivative to zero to find critical points:\[ \frac{1 - \ln(x)}{x^2} = 0. \]This implies that the numerator must be zero, so solve \( 1 - \ln(x) = 0 \), which gives \( \ln(x) = 1 \). Therefore, \( x = e \) is the critical point.

Step 3: Determine the intervals of increase and decrease

Consider the sign of \( f'(x) = \frac{1 - \ln(x)}{x^2} \) to determine the behavior of the function on different intervals:- For \( x < e \): If \( x < e \), then \( \ln(x) < 1 \) which makes \( 1 - \ln(x) > 0 \). Thus, \( f'(x) > 0 \), meaning \( f(x) \) is increasing.- For \( x > e \): If \( x > e \), then \( \ln(x) > 1 \) which makes \( 1 - \ln(x) < 0 \). Thus, \( f'(x) < 0 \), meaning \( f(x) \) is decreasing.

Step 4: Apply the First Derivative Test at critical point x = e

Since \( f(x) \) switches from increasing to decreasing as \( x \) passes through \( e \), by the First Derivative Test, \( f(e) \) is a local maximum. This is because the derivative changes from positive to negative at \( x = e \).

Key Concepts

Increasing and Decreasing Intervals

To determine on which intervals a function is increasing or decreasing, we first need to find its first derivative. The first derivative, denoted as \( f'(x) \), tells us about the slope of the tangent line to the curve at any point \( x \). When \( f'(x) > 0 \), the function \( f(x) \) is increasing in that interval. On the other hand, when \( f'(x) < 0 \), the function is decreasing.

For the function \( f(x) = \ln(x)/x \), we've found that its derivative \( f'(x) \) is \( \frac{1 - \ln(x)}{x^2} \). Here's how you determine the behavior of this function based on \( f'(x) \):

• When \( x < e \), the expression \( \ln(x) < 1 \) ensures \( 1 - \ln(x) > 0 \), making \( f'(x) > 0 \), so \( f(x) \) is increasing.
• Conversely, when \( x > e \), the expression \( \ln(x) > 1 \) means \( 1 - \ln(x) < 0 \), making \( f'(x) < 0 \), so \( f(x) \) is decreasing.

To apply this knowledge, always consider the sign of the first derivative over the intervals to find where the function is increasing or decreasing.

Critical Points

Critical points of a function occur where its first derivative is zero or undefined. These points are important because they might represent local maximums, local minimums, or points of inflection.

For the function \( f(x) = \ln(x)/x \), the critical points are found by setting the first derivative equal to zero: \( f'(x) = \frac{1 - \ln(x)}{x^2} = 0 \). This equation simplifies to \( 1 - \ln(x) = 0 \), which further simplifies to \( \ln(x) = 1 \). Solving this, we find that \( x = e \) is a critical point.

Critical points help us identify regions where the function might change its behavior. After identifying these points, we further analyze the function to determine its local maxima and minima.

Quotient Rule

When dealing with functions that are quotients of two differentiable functions, such as \( f(x) = \frac{u}{v} \), the quotient rule is a crucial tool. This rule provides a formula for finding the derivative of such functions:

The quotient rule states:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]

For our specific function \( f(x) = \ln(x)/x \), we identify \( u = \ln(x) \) and \( v = x \). Calculating the derivatives, \( u' = \frac{1}{x} \) and \( v' = 1 \). Plugging these into the quotient rule gives us the first derivative:
\[ f'(x) = \frac{\frac{1}{x} \cdot x - \ln(x) \cdot 1}{x^2} = \frac{1 - \ln(x)}{x^2} \]

This derivative helps break down the behavior of the original function and analyze its intervals of increase and decrease.

Local Maximum and Minimum

Local maximum and minimum points of a function are points where the function reaches a peak (maximum) or a trough (minimum) within a certain interval. We use the First Derivative Test to determine these points. When applied, if the derivative changes from positive to negative at a critical point, that point is a local maximum. If it changes from negative to positive, it's a local minimum.

Consider the critical point \( x = e \) for our function \( f(x) = \ln(x)/x \).

• For \( x < e \), \( f'(x) > 0 \) indicates \( f(x) \) is increasing.
• For \( x > e \), \( f'(x) < 0 \) implies \( f(x) \) is decreasing.

Since \( f'(x) \) changes from positive to negative as \( x \) passes through \( e \), by the First Derivative Test, the point \( f(e) \) is a local maximum point. This test helps pinpoint where the function achieves local highs and lows within specified intervals.