Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Max at \( x = e - 1 \) is \( \frac{1}{e} \); Min at \( x = 0 \) is 0.

Step 1: Find the first derivative

To find the critical points of the function, we need to take the derivative and set it to zero. Given \( g(x)=\frac{\ln(x+1)}{x+1} \), we use the quotient rule to find the derivative:\(g'(x)= \frac{(x+1)\cdot\frac{1}{x+1} - \ln(x+1)\cdot 1}{(x+1)^2} = \frac{1 - \ln(x+1)}{(x+1)^2}.\)

Step 2: Set the first derivative to zero

We set the first derivative equal to zero to find the critical points:\(\frac{1 - \ln(x+1)}{(x+1)^2} = 0.\)This simplifies to solving:\(1 - \ln(x+1) = 0 \implies \ln(x+1) = 1.\)Taking the exponential of both sides, we have:\(x+1 = e^1 \implies x = e - 1.\)

Step 3: Confirm critical points are within the interval

Check if the critical point \(x = e - 1\) lies within the interval \([0, 3]\):Since \(e \approx 2.718\), we find that \(x = e - 1 \approx 1.718\), which is indeed between 0 and 3.

Step 4: Evaluate the function at critical points and endpoints

Evaluate \(g(x)\) at the critical point and the endpoints of the interval.- At \(x = 0\): \(g(0) = \frac{\ln(0+1)}{0+1} = 0.\)- At \(x = e - 1\): \(g(e-1) = \frac{\ln(e)}{e} = \frac{1}{e}.\)- At \(x = 3\): \(g(3) = \frac{\ln(3+1)}{3+1} = \frac{\ln(4)}{4}.\)

Step 5: Determine the maximum and minimum values

Compare the function values at each point:- \(g(0) = 0\)- \(g(e-1) = \frac{1}{e} \approx 0.368\)- \(g(3) = \frac{\ln(4)}{4} \approx 0.3466\)The maximum value is \(\frac{1}{e} \approx 0.368\) at \(x = e - 1\), and the minimum value is 0 at \(x = 0\).

Key Concepts

First Derivative

To find critical points in a function, we first need to calculate its derivative. The derivative gives us an insight into the slope of the tangent line to the curve of the function at any given point.
For the function \( g(x) = \frac{\ln(x+1)}{x+1} \), we specifically find its first derivative. This is crucial because setting the derivative to zero helps us locate the potential turning points in the function's behavior. These turning points are areas where the function may reach a maximum or minimum value.
Computing the first derivative involves using derivative rules effectively. In this exercise, the derivative is derived as \( g'(x) = \frac{1 - \ln(x+1)}{(x+1)^2} \). This expression will guide us further into finding out where the critical points are by solving the equation \( g'(x) = 0 \).

Quotient Rule

The quotient rule is a method we use in calculus to differentiate functions that are presented as fractions. When a function \( y = \frac{u}{v} \) involves a numerator \( u \) and a denominator \( v \), the derivative is obtained using the formula:

• \( \frac{du}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

In the case of \( g(x) = \frac{\ln(x+1)}{x+1} \), the numerator \( u \) is \( \ln(x+1) \) and the denominator \( v \) is \( x+1 \).
This rule allows us to handle the complexities of dividing functions, and it provides the framework needed for finding the derivative of \( g(x) \).

Maximum Value

Critical points found by setting the first derivative to zero can either be maximum, minimum, or saddle points. In the problem, we evaluated the function at its critical points and endpoints to find these values. The critical point \( x = e - 1 \) falls within our interested interval \( [0, 3] \).
By calculating \( g(e-1) \), we determined it equals \( \frac{1}{e} \) which is approximately 0.368.
This is compared against calculations at endpoint values like \( x = 0 \) and \( x = 3 \). Among these computational results, the value \( \frac{1}{e} \approx 0.368 \) at \( x = e-1 \) is the highest, thus indicating the point of maximum value for this function on the given interval.

Minimum Value

Similarly, understanding the notion of minimum value involves comparing the values of a function at various points, particularly at the critical and boundary points of the given interval.
For \( g(x) \), we determined a critical point \( x = 0 \), yielding a function value of 0. This is confirmed by calculating \( g(0) = 0 \) since both the numerator and the denominator are equal at this point.
This evaluation, along with those at other specified points, showcases the minimum value in the interval \( [0, 3] \) to be at \( x = 0 \). As no other point yielded a lower value than 0, this becomes the minimum value for our function.