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Chapter 3: Problem 26

Locate the absolute extrema of the function on the closed interval. $$ y=x \ln (x+3),[0,3] $$

Short Answer

Expert verified
The absolute maximum of the function over the interval [0,3] is \(3 \ln(6)\) at \(x=3\), and the absolute minimum is \(0\) at \(x=0\).

Step by step solution

01

Find the Derivative

The first step is to find the derivative of the function \(y = x \ln (x + 3)\). Using the product rule (where the derivative of \(u \cdot v\) is \(u'v + uv'\)), let \(u = x\) and \(v = \ln(x+3)\). Thus, \(u' = 1\) and \(v' = \frac{1}{x+3}\), and the derivative \(y' = (1)\ln(x + 3) + (x) \cdot \frac{1}{x+3} =\ln(x+3) + \frac{x}{x+3}\).
02

Find the Critical Points

The critical points are where the derivative is either zero or undefined. Set the derivative equal to zero and solve for \(x\). After solving \(x+3-\frac{x^2+3x}{x+3} = 0\), we get one critical point, \(x = 1\). We also see that the derivative is undefined at \(x = -3\), but this is not in our interval.
03

Evaluate the Endpoints and Critical Point

Now evaluate the function \(y = x \ln (x + 3)\) at the endpoints of the interval and at the critical point. We have \(y(0) = 0\), \(y(3) = 3 \ln (6)\), and \(y(1) = 1 \ln (4)\)
04

Determine the Absolute Extrema

The function will have its absolute maximum and minimum at either the endpoints or the critical point. So, we just compare these three values. We observe that \(0 \leq \ln(4) \leq 3 \ln(6)\). Thus the absolute minimum is \(0\) at \(x=0\) and the maximum is \(3 \ln(6)\) at \(x=3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
The derivative of a function gives us the rate at which the function's value changes with respect to changes in its input. It's crucial for determining the slope of the function at any given point. In our exercise, we begin with the function \( y = x \ln (x+3) \). To find the derivative, we need to apply the product rule. This will help us in further steps to determine critical points and evaluate endpoints for extrema.
Product Rule
When dealing with differentiating products of two functions, the product rule is our best ally. The product rule states that if we have two functions \( u \) and \( v \), then the derivative of their product is given by \( u'v + uv' \). In our function, setting \( u = x \) and \( v = \ln(x+3) \) is strategic. Here, \( u' = 1 \) as the derivative of \( x \) is simply 1. The derivative of \( v \), or \( v' \), is \( \frac{1}{x+3} \). Using the product rule, the derivative becomes a sum of products: \( \ln(x+3) \) plus \( \frac{x}{x+3} \). This form is much easier to work with in determining critical points later on.
Critical Points
Critical points are the values of \( x \) where the derivative is either zero or undefined. These points are important because they give us potential locations of local extrema 鈥� spots where the function might reach a maximum or minimum value. In our step-by-step solution, we set the derivative \( y' = \ln(x+3) + \frac{x}{x+3} \) equal to zero to find critical points. Solving this involves setting the numerator of \( \frac{x^2+3x-x-3x}{(x+3)} = 0 \), simplifying, and rearranging to find \( x = 1 \). Always verify that these points lie within the given interval before using them in further calculations.
Endpoints Evaluation
After identifying the critical points, the next step is evaluating the function at these points and at the endpoints of the interval. Endpoints can often provide absolute extrema, especially in closed intervals. In our given interval \([0,3]\), the function needs assessing at \( x=0 \), \( x=3 \), and the critical point \( x=1 \). Evaluating gives \( y(0) = 0 \), \( y(3) = 3 \ln(6) \), and \( y(1) = 1 \ln(4) \). We compare these values to determine the absolute minimum and maximum of the function in this interval. This approach ensures we account for all possibilities of extrema by not only relying on critical points but also examining the behavior at the interval's boundaries.

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Most popular questions from this chapter

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