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Chapter 2: Problem 21

Estimate the instantaneous rate of change of the function \(f(x)=x \ln x\) at \(x=1\) and at \(x=2 .\) What do these values suggest about the concavity of the graph between 1 and \(2 ?\)

Short Answer

Expert verified
At \(x=1\), the rate of change is 1, and at \(x=2\), it is approximately 1.693. The graph is concave up between 1 and 2.

Step by step solution

01

Find the Derivative

The first step in estimating the instantaneous rate of change of the function is to find its derivative. The derivative of a function gives the rate at which the function is changing at any point on the graph.For the function \(f(x) = x \ln x\), we use the product rule, which states that the derivative of a product \(uv\) is \(u'v + uv'\). Here, \(u = x\) and \(v = \ln x\). Thus,\[ \frac{d}{dx}[x \ln x] = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1. \]
02

Calculate the Derivative at x=1

Now we find the derivative of the function at \(x=1\) using the expression derived in Step 1.\[ f'(1) = \ln(1) + 1 = 0 + 1 = 1. \]
03

Calculate the Derivative at x=2

Substitute \(x=2\) into the derivative to find the instantaneous rate of change at this point.\[ f'(2) = \ln(2) + 1. \]Using the known value \(\ln(2) \approx 0.693\), we get\[ f'(2) \approx 0.693 + 1 = 1.693. \]
04

Analyze Concavity

To determine the concavity between \(x = 1\) and \(x = 2\), we look at the sign of the second derivative in that interval. Calculate the second derivative by differentiating the first derivative, \(f'(x) = \ln x + 1\).\[ f''(x) = \frac{d}{dx}(\ln x + 1) = \frac{1}{x}. \]Since \(f''(x) = \frac{1}{x}\) is positive for \(x > 0\), the function \(f(x) = x \ln x\) is concave up on the interval from \(1\) to \(2\).

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Key Concepts

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

Derivative
When we talk about the instantaneous rate of change of a function, what we're referring to is the derivative. The derivative of a function at a particular point tells us how quickly the function's value is changing at that exact point.
For the function given in the exercise, \( f(x) = x \ln x \), we are interested in finding this rate at certain points, specifically at \( x = 1 \) and \( x = 2 \).
To find the derivative, we employ a fundamental tool in calculus called the product rule.
  • The product rule is used when you need to differentiate a product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), and their product is \( uv \), then the derivative is given by \( u'v + uv' \).
  • In our case, \( u(x) = x \) and \( v(x) = \ln x \), so applying the product rule results in \( f'(x) = \ln x + 1 \).
Once we have our derivative, calculating it at particular points (like \( x = 1 \) or \( x = 2 \)) gives us the instantaneous rates of change at those points.
Concavity
Concavity helps us understand the direction in which a curve is bending. Whether a function is bending upwards or downwards is determined by studying its second derivative. In these terms:
  • If the second derivative, \( f''(x) \), is positive over an interval, the function is said to be concave up, resembling a U-shape.
  • If \( f''(x) \) is negative, the function is concave down, resembling an upside-down U.
For our function, the second derivative is \( f''(x) = \frac{1}{x} \), which is always positive for \( x > 0 \).
Therefore, between the points \( x = 1 \) and \( x = 2 \), the function \( f(x) = x \ln x \) is concave up. This means the graph is bending upwards along this interval.
Product Rule
The product rule is essential when differentiating products of two functions. Here's a breakdown of how and when to use it:
  • It's crucial whenever you're faced with a differentiation task involving two functions multiplied together. Simply doing the derivative of each part separately won't work.
  • Think of it like this: if you have a product \( uv \), the derivative becomes \( u'v + uv' \). This means you differentiate \( u \), keep \( v \) as it is, and vice versa, then sum the two results.
In the context of the example \( f(x) = x \ln x \), the product rule was used to effectively split the work between the two parts, \( x \) and \( \ln x \).
Applying the rule ensured that we correctly derived the expression \( \ln x + 1 \) for the first derivative, capturing how both the \( x \) and the \( \ln x \) components contribute to the change at any given point.

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Most popular questions from this chapter

Suppose \(C(r)\) is the total cost of paying off a car loan borrowed at an annual interest rate of \(r \% .\) What are the units of \(C^{\prime}(r) ?\) What is the practical meaning of \(C^{\prime}(r) ?\) What is its sign?

In \(1913,\) Carlson \(^{28}\) conducted the classic experiment in which he grew yeast, Saccharomyces cerevisiae, in laboratory cultures and collected data every hour for 18 hours. Table 2.12 shows the yeast population, \(P,\) at representative times \(t\) in hours. (a) Calculate the average rate of change of \(P\) per hour for the time intervals shown between 0 and 18 hours. (b) What can you say about the sign of \(d^{2} P / d t^{2}\) during the period \(0-18\) hours? $$\begin{array}{c|r|r|r|r|r|r|r|r|r|r}\hline t & 0 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 \\ \hline P & 9.6 & 29.0 & 71.1 & 174.6 & 350.7 & 513.3 & 594.8 & 640.8 & 655.9 & 661.8 \\ \hline\end{array}$$

An industrial production process costs \(C(q)\) million dollars to produce \(q\) million units; these units then sell for \(R(q)\) million dollars. If \(C(2.1)=5.1, R(2.1)=6.9\) \(M C(2.1)=0.6,\) and \(M R(2.1)=0.7,\) calculate (a) The profit earned by producing 2.1 million units (b) The approximate change in revenue if production increases from 2.1 to 2.14 million units. (c) The approximate change in revenue if production decreases from 2.1 to 2.05 million units. (d) The approximate change in profit in parts (b) and (c).

A company's cost of producing \(q\) liters of a chemical is \(C(q)\) dollars; this quantity can be sold for \(R(q)\) dollars. Suppose \(C(2000)=5930\) and \(R(2000)=7780\) (a) What is the profit at a production level of \(2000 ?\) (b) If \(M C(2000)=2.1\) and \(M R(2000)=2.5,\) what is the approximate change in profit if \(q\) is increased from 2000 to \(2001 ?\) Should the company increase or decrease production from \(q=2000 ?\) (c) If \(M C(2000)=4.77\) and \(M R(2000)=4.32\) should the company increase or decrease production from \(q=2000 ?\)

When you breathe, a muscle (called the diaphragm) reduces the pressure around your lungs and they expand to fill with air. The table shows the volume of a lung as a function of the reduction in pressure from the diaphragm. Pulmonologists (lung doctors) define the compliance of the lung as the derivative of this function. 10 (a) What are the units of compliance? (b) Estimate the maximum compliance of the lung. (c) Explain why the compliance gets small when the lung is nearly full (around 1 liter). $$\begin{array}{|c|c|} \hline \begin{array}{c} \text { Pressure reduction } \\ \text { (cm of water) } \end{array} & \begin{array}{c} \text { Volume } \\ \text { (iters) } \end{array} \\ \hline 0 & 0.20 \\ \hline 5 & 0.29 \\ \hline 10 & 0.49 \\ \hline 15 & 0.70 \\ \hline 20 & 0.86 \\ \hline 25 & 0.95 \\ \hline 30 & 1.00 \\ \hline \end{array}$$
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