Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-26, identify the critical points and find the maximum value and minimum value on the given interval. $$ g(x)=\frac{\ln (x+1)}{x+1} ; I=[0,3] $$

Step-by-Step Solution

Answer

Maximum value is 0.368 at \( x = e-1 \); minimum value is 0 at \( x = 0 \).

Step 1: Differentiate the function

To find the critical points, we first need to differentiate the function. The given function is \( g(x) = \frac{\ln(x+1)}{x+1} \). Let \( u = \ln(x+1) \) and \( v = x+1 \). Then, \( g(x) = \frac{u}{v} \), and we use the quotient rule \( \left( \frac{u}{v} \right)' = \frac{v u' - u v'}{v^2} \). Differentiating, \( u' = \frac{1}{x+1} \) and \( v' = 1 \), so \( g'(x) = \frac{(x+1) \cdot \frac{1}{x+1} - \ln(x+1) \cdot 1}{(x+1)^2} = \frac{1 - \ln(x+1)}{(x+1)^2} \).

Step 2: Find Critical Points

Critical points occur where \( g'(x) = 0 \) or where \( g'(x) \) is undefined. Solving \( g'(x) = 0 \), we have \( \frac{1 - \ln(x+1)}{(x+1)^2} = 0 \). This simplifies to \( 1 - \ln(x+1) = 0 \) or \( \ln(x+1) = 1 \) which gives \( x+1 = e \), meaning \( x = e - 1 \). Since \( e \approx 2.718 \), \( x = e - 1 \approx 1.718 \), which is within the interval \([0, 3]\). Check that \( g'(x) \) is defined in this interval because \( x+1 > 0 \) for all \( x \) in \([0, 3]\).

Step 3: Evaluate the function at endpoints and critical points

Evaluate \( g(x) \) at the critical point \( x = e - 1 \), as well as at the endpoints \( x = 0 \) and \( x = 3 \). We have: \[ g(0) = \frac{\ln(0+1)}{0+1} = \frac{0}{1} = 0 \] \[ g(e-1) = \frac{\ln(e)}{e} = \frac{1}{e} \approx 0.368 \] \[ g(3) = \frac{\ln(4)}{4} \approx \frac{1.386}{4} \approx 0.3465 \].

Step 4: Identify maximum and minimum values

By comparing the evaluated values: - \( g(0) = 0 \)- \( g(e-1) \approx 0.368 \)- \( g(3) \approx 0.3465 \)The maximum value of \( g(x) \) on \([0, 3]\) is \( 0.368 \) at \( x = e-1 \), and the minimum value is \( 0 \) at \( x = 0 \).

Key Concepts

Quotient Rule

The quotient rule is a handy tool in calculus used to differentiate functions that are ratios of two differentiable functions. In mathematical terms, if you have a function \( g(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable, the derivative \( g'(x) \) is given by:

• \( g'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \)

This means we differentiate the numerator and the denominator separately, then apply this formula.
In our example with \( g(x) = \frac{\ln(x+1)}{x+1} \), \( u = \ln(x+1) \) and \( v = x+1 \).
When solved, we find the derivative using:

• \( u'(x) = \frac{1}{x+1} \)
• \( v'(x) = 1 \)

Applying the quotient rule, we find \( g'(x) = \frac{1 - \ln(x+1)}{(x+1)^2} \).
This derivative helps us find critical points.

Differentiation

Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. It essentially gives the rate at which a function is changing at any point, which is crucial in identifying critical points.
In our problem, differentiating \( g(x) = \frac{\ln(x+1)}{x+1} \) helps us understand where the function increases or decreases and its turning points.
To be specific, differentiation allows us to apply rules such as the quotient rule to decompose complex functions.
By finding where the derivative \( g'(x) = 0 \), we locate the x-values where the function might switch direction, which are potential critical points or extremum points.
This understanding enables us to analyze the function more deeply.

Extreme Value Theorem

The Extreme Value Theorem (EVT) tells us that a continuous function over a closed interval \([a, b]\) will have both a maximum and a minimum value.
In essence, the theorem guarantees that these extreme values are attained either at critical points within the interval or at the interval's endpoints.
In our example, we looked at \( g(x) \) over the interval \([0, 3]\). Using EVT, we evaluated the function at critical points and endpoints:

• Critical Point: \( x = e - 1 \)
• Endpoints: \( x = 0 \) and \( x = 3 \)

By doing this, we determined that the maximum occurred at the critical point, and the minimum occurred at one of the endpoints.
This methodical approach guarantees that we've identified all potential extremum values, ensuring a comprehensive evaluation.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a special type of logarithm with the base \( e \), where \( e \approx 2.718 \). It appears naturally in many areas of mathematics, especially in calculus when working with exponential growth and decay.
The derivative of \( \ln(x) \) is particularly simple and useful: it's \( \frac{1}{x} \).
In the given function \( g(x) = \frac{\ln(x+1)}{x+1} \), the natural logarithm helps form the numerator.
When differentiating \( \ln(x+1) \), we get a straightforward derivative of \( \frac{1}{x+1} \), indicating smooth growth rates relative to the input increases.
This insight allows us to apply the quotient rule efficiently and affects how the entire function behaves over the defined interval, contributing to finding critical points.