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Chapter 4: Problem 26

Find the exact global maximum and minimum values of the function. The domain is all real numbers unless otherwise specified. $$ f(x)=x-\ln x \text { for } x>0 $$

Short Answer

Expert verified
The global minimum value is 1 at \( x = 1 \); there is no global maximum.

Step by step solution

01

Identify the Function and its Domain

We are given the function \( f(x) = x - \ln x \) with the domain \( x > 0 \). Our task is to find its global maximum and minimum values in this domain.
02

Find the Critical Points

To find the critical points, we first need to find the derivative of \( f(x) \). The derivative is \( f'(x) = 1 - \frac{1}{x} \). Setting this equal to zero gives \( 1 - \frac{1}{x} = 0 \). Solving this equation, we find the critical point \( x = 1 \).
03

Analyze the Critical Points

To determine if this critical point is a maximum or minimum, we examine the second derivative. Calculating the second derivative yields \( f''(x) = \frac{1}{x^2} \). Evaluating this at the critical point gives \( f''(1) = 1 \), which is positive, indicating a local minimum at \( x = 1 \).
04

Evaluate Function Extrema

Since there are no endpoints in the given domain \( x > 0 \), we will additionally evaluate the behavior of \( f(x) \) as \( x \to \infty \) and as \( x \to 0^+ \). As \( x \to \infty \), \( f(x) = x - \ln x \to \infty \). As \( x \to 0^+ \), \( \ln x \to -\infty \), making \( f(x) \to \infty \). Hence, there is no global minimum at the boundaries, but the global maximum is not definitively shown at just \( x \to \infty \) or \( x \to 0^+ \).
05

Conclusion

Since the analysis shows that \( f(x) \) increases beyond the local minimum at \( x = 1 \) as \( x \to \infty \) and \( x \to 0^+ \), it indicates there is no maximum in the strict sense but it decreases initially achieving a local minimum at \( x = 1 \). Therefore, the global minimum value is \( f(1) = 1 - \ln(1) = 1 \), while the function does not attain a global maximum within the constraints, as it continues to increase indefinitely.

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Key Concepts

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

Critical Points
Finding critical points is the first step in determining the global extrema of a function. Critical points occur where the derivative of the function is zero or undefined. For the function \[ f(x) = x - \ln x \] the derivative is calculated as:\[ f'(x) = 1 - \frac{1}{x} \]Setting the derivative to zero, \[ 1 - \frac{1}{x} = 0 \]results in\[ \frac{1}{x} = 1 \] Solving this equation for \( x \), we find a critical point at \( x = 1 \).- Critical points are potential locations of local maxima and minima.- They help us understand the behavior of the function at particular points in the domain.
Once we have the critical points, the next step is to use the second derivative test to determine whether these points are maxima, minima, or points of inflection.
Second Derivative Test
After finding the critical points, the second derivative test helps determine the nature of these points. The second derivative test involves taking the second derivative of the function and evaluating it at the critical points.For our function,\[ f''(x) = \frac{1}{x^2} \]Evaluating at the critical point \( x = 1 \), we find:\[ f''(1) = \frac{1}{1^2} = 1 \]A positive second derivative means the function is concave up at that point, indicating a local minimum at \( x = 1 \). Conversely, if the second derivative were negative, the function would be concave down, indicating a local maximum.

- Use the second derivative to confirm the nature of critical points- Positive results indicate minima, negative results indicate maximaKnowing the nature of these critical points allows us to understand the regions where global maxima or minima might occur.
Natural Logarithm
The natural logarithm, denoted as \( \ln x \), is the power to which the base \( e \) (approximately 2.71828) must be raised to yield the number \( x \). It is an important mathematical function with properties that are useful in calculus and analysis.Understanding how \( \ln x \) behaves helps when solving problems involving derivatives or integrals:
  • As \( x \to \infty \), \( \ln x \to \infty \), meaning it grows very slowly compared to polynomial functions.

  • As \( x \to 0^+ \), \( \ln x \to -\infty \), indicating a steep negative slope approaching \( x = 0 \).
  • The derivative of \( \ln x \) is \( \frac{1}{x} \), which decreases as \( x \) increases.

The natural logarithm鈥檚 characteristics are crucial when we analyze functions like \( x - \ln x \), as they influence both the position of critical points and the method of determining extrema.

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Most popular questions from this chapter

During a flu outbreak in a school of 763 children, the number of infected children, \(I\), was expressed in terms of the number of susceptible (but still healthy) children, \(S\), by the expression $$I=192 \ln \left(\frac{S}{762}\right)-S+763 .$$ What is the maximum possible number of infected children?

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. \(f(x)=3 x^{5}-5 x^{3}\)

A rumor spreads among a group of 400 people. The number of people, \(N(t)\), who have heard the rumor by time \(t\) in hours since the rumor started to spread can be approximated by a function of the form $$ N(t)=\frac{400}{1+399 e^{-0.4 t}} $$ (a) Find \(N(0)\) and interpret it. (b) How many people will have heard the rumor after 2 hours? After 10 hours? (c) Graph \(N(t)\). (d) Approximately how long will it take until half the people have heard the rumor? Virtually everyone? (e) Approximately when is the rumor spreading fastest?

The income elasticity of demand for a product is defined as \(E_{\text {income }}=|I / q \cdot d q / d I|\) where \(q\) is the quantity demanded as a function of the income \(I\) of the consumer. What does \(E_{\text {income }}\) tell you about the sensitivity of the quantity of the product purchased to changes in the income of the consumer?

The quantity of a drug in the bloodstream \(t\) hours after a tablet is swallowed is given, in \(\mathrm{mg}\), by $$q(t)=20\left(e^{-t}-e^{-2 t}\right)$$ (a) How much of the drug is in the bloodstream at time \(t=0 ?\) (b) When is the maximum quantity of drug in the bloodstream? What is that maximum? (c) In the long run, what happens to the quantity?
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