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Chapter 4: Problem 30

Find: (a) the intervals on which \(f\) is increasing, (b) the intervals on which \(f\) is decreasing, (c) the open intervals on which \(f\) is concave up, (d) the open intervals on which \(f\) is concave down, and (e) the \(x\) -coordinates of all inflection points. $$f(x)=x^{3} \ln x$$

Short Answer

Expert verified
(a) Increasing on \((e^{-\frac{1}{3}}, \infty)\); (b) Decreasing on \((0, e^{-\frac{1}{3}})\).

Step by step solution

01

Find the First Derivative

To determine where the function is increasing or decreasing, first calculate the first derivative of \(f(x)\). Apply the product rule: if \(f(x) = u(x)v(x)\), then \(f'(x) = u'(x)v(x) + u(x)v'(x)\). Here, let \(u(x) = x^3\) and \(v(x) = \ln x\). Then, \(u'(x) = 3x^2\) and \(v'(x) = \frac{1}{x}\). The first derivative is:\[ f'(x) = 3x^2 \ln x + x^2 \]
02

Find Critical Points

Set \(f'(x) = 0\) to find critical points. We have:\[ 3x^2 \ln x + x^2 = x^2(3 \ln x + 1) = 0 \]This gives us \(x^2 = 0\) or \(3 \ln x + 1 = 0\). \(x^2 = 0\) gives us \(x = 0\). Solving \(3 \ln x + 1 = 0\) gives \(x = e^{-\frac{1}{3}}\). The critical points are \(x = 0\) and \(x = e^{-\frac{1}{3}}\). Since the domain of \(f(x)\) is \((0, \infty)\), we disregard \(x = 0\).
03

Determine Intervals of Increase/Decrease

Use the critical point \(x = e^{-\frac{1}{3}}\) to test intervals on \(f'(x)\). Choose test points in the intervals \((0, e^{-\frac{1}{3}})\) and \((e^{-\frac{1}{3}}, \infty)\):- For \(x \in (0, e^{-\frac{1}{3}})\), pick \(x = e^{-1}\) and test sign of \(3x^2 \ln x + x^2\).\- For \(x \in (e^{-\frac{1}{3}}, \infty)\), pick \(x = 1\).- For \(x = e^{-1}\), since \(\ln\left(e^{-1}\right) = -1\), \(3(-1) + 1 = -2 < 0\), so \(f'(x) < 0\) (decreasing).- For \(x = 1\), \(\ln 1 = 0\), \(3(0) + 1 = 1 > 0\), so \(f'(x) > 0\) (increasing).

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Key Concepts

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

First Derivative
The first derivative is a fundamental concept in calculus. It gives us a formula to find the rate at which a function is changing at any given point. To find the first derivative of the function \( f(x) = x^3 \ln x \), we can use the product rule.
  • The product rule states that if you have a product of two functions, \( u(x) \) and \( v(x) \), then the derivative is \( u'(x) v(x) + u(x) v'(x) \).
  • For \( f(x) = x^3 \ln x \), imagine \( u(x) = x^3 \) and \( v(x) = \ln x \). The derivatives are \( u'(x) = 3x^2 \) and \( v'(x) = \frac{1}{x} \).
  • Plug these into the product rule: \( f'(x) = 3x^2 \ln x + x^2 \).
This derivative tells us how \( f(x) \) changes as \( x \) changes.
Increasing and Decreasing Intervals
Once we have the first derivative of a function, we can use it to determine where the function is increasing or decreasing. A function is increasing where its derivative is positive, and decreasing where its derivative is negative.
  • Look at the first derivative \( f'(x) = 3x^2 \ln x + x^2 \).
  • Set \( f'(x) = 0 \) to find critical points, which can help break the number line into intervals.
In this example, after setting \( f'(x) = 0 \), we find critical points at \( x = e^{-\frac{1}{3}} \). Use test points in intervals \((0, e^{-\frac{1}{3}})\) and \((e^{-\frac{1}{3}}, \infty)\) to determine sign of \( f'(x) \). The function decreases on \( (0, e^{-\frac{1}{3}}) \) and increases on \( (e^{-\frac{1}{3}}, \infty) \).
Critical Points
Critical points are where the function's derivative equals zero or does not exist, showing potential changes in the direction of the function. To find critical points:
  • Take the derivative \( f'(x) = 3x^2 \ln x + x^2 \) and set it equal to zero.
  • Solve: \( x^2(3 \ln x + 1) = 0 \).
This yields \( x^2 = 0 \) or \( 3 \ln x + 1 = 0 \). Discard \( x = 0 \) as it is not in the domain of \( \ln x \). The critical point within the domain is \( x = e^{-\frac{1}{3}} \). This is where potential changes from increasing to decreasing or vice versa occur.
Concavity
Concavity describes the direction of the curve. If a function is concave up, it looks like a cup; if concave down, it looks like a cap. We must find the second derivative to determine concavity.
  • Compute the second derivative of \( f(x) \).
  • Analyze \( f''(x) \) to determine intervals where the function is concave up (where \( f''(x) > 0 \)) or concave down (where \( f''(x) < 0 \)).
Finding these intervals helps us understand more about the curvature of the function and identify potential points where this curvature changes.
Inflection Point
An inflection point is where a function's graph changes concavity. It occurs where the second derivative equals zero or is undefined, and the sign of the second derivative changes at this point.
  • After finding the second derivative, \( f''(x) \), solve \( f''(x) = 0 \).
  • Verify that there is a change in sign of \( f''(x) \) around the solution to confirm it as an inflection point.
Inflection points provide valuable insight into a function's behavior, indicating where it transitions between concave up and concave down. Identifying these points give a clearer picture of how the graph looks over different intervals.

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Most popular questions from this chapter

Use a graphing utility to make a conjecture about the relative extrema of \(f,\) and then check your conjecture using either the first or second derivative test. $$f(x)=\frac{2}{e^{x}+e^{-x}}$$

(a) Show that if \(f\) and \(g\) are functions for which \(f^{\prime}(x)=g(x) \quad\) and \(\quad g^{\prime}(x)=f(x)\) for all \(x,\) then \(f^{2}(x)-g^{2}(x)\) is a constant. (b) Show that the function \(f(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)\) and the function \(g(x)=\frac{1}{2}\left(e^{x}-e^{-x}\right)\) have this property.

(a) Prove that a general cubic polynomial $$ f(x)=a x^{3}+b x^{2}+c x+d \quad(a \neq 0) $$ has exactly one inflection point. (b) Prove that if a cubic polynomial has three \(x\) -intercepts, then the inflection point occurs at the average value of the intercepts. (c) Use the result in part (b) to find the inflection point of the cubic polynomial \(f(x)=x^{3}-3 x^{2}+2 x,\) and check your result by using \(f^{\prime \prime}\) to determine where \(f\) is concave up dow

Discuss the relative advantages or disadvantages of using the first derivative test versus using the second derivative test to classify candidates for relative extrema on the interior of the domain of a function. Include specific examples to illustrate your points.

A rectangular plot of land is to be fenced off so that the area enclosed will be \(400 \mathrm{ft}^{2},\) Let \(L\) be the length of fencing needed and \(x\) the length of one side of the rectangle. Show that \(L=2 x+800 / x\) for \(x>0,\) and sketch the graph of \(L\) versus \(x\) for \(x>0\).
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