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Chapter 4: Problem 4

In Exercises \(1-6,\) use the First Derivative Test to determine the local extreme values of the function, and identify any absolute extrema. Support your answers graphically. $$y=x e^{1 / x}$$

Short Answer

Expert verified
The local maximum is at the point \((-1, - e^{-1})\) and the local minimum is at the point \((1, e^{-1})\). The function does not have any points of absolute extrema.

Step by step solution

01

Compute the First Derivative

Differentiate the given function \(y = xe^{1/x}\) using the product rule for derivatives which states that the derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. This yields \[ y' = e^{1/x} - xe^{1/x}/x^2 \]
02

Find Critical Points

Setting the derivative equal to zero allows us to find the critical points that are candidates for local extreme values. Thus, \[ e^{1/x} - xe^{1/x}/x^2 = 0 \] Solving this equation gives \( x = 1 \) and \( x = -1 \).
03

Apply the First Derivative Test

Plugging in values that are less than -1, between -1 and 1, and greater than 1 into the derivative equation tests the sign. A change from positive to negative indicates a local maximum, while a change from negative to positive indicates a local minimum. Here, the function decreases in \((-鈭�, -1)\) and \((-1, 1)\) and increases in \((1, +鈭�)\). Hence, \(x=-1\) is a local maximum and \(x=1\) is a local minimum.
04

Find the Local Extreme Values

Substitute \( x = -1 \) and \( x = 1 \) into the function to find the local extreme values. The local maximum is \((-1, -e^{-1})\) and the local minimum is \((1, e^{-1})\)
05

Identify the Absolute Extrema

Because the function decreases without bound as \( x \) approaches negative infinity and increases without bound as \( x \) approaches positive infinity, it does not have any absolute extrema.

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Key Concepts

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

Local Extrema
Local extrema of a function refer to local maximum or minimum points within a specified interval. These are the highest or lowest points in a small neighborhood around the point, although they might not be the highest or lowest points overall for the function.

In the First Derivative Test, finding local extrema involves analyzing the changes in the sign of the first derivative around critical points.
  • If the derivative changes from positive to negative, the function has a local maximum at that point.
  • Conversely, if the derivative changes from negative to positive, there is a local minimum.
This test helps determine whether a critical point is a local maximum, local minimum, or neither. In our example, we found that at points \(x = -1\) and \(x = 1\), the function has local extrema.
Critical Points
Critical points are points on the graph of a function where the first derivative is zero or undefined. Identifying these points is crucial for locating potential local extrema. They are often where the graph changes direction, but not always.

To find critical points, we set the derivative equal to zero and solve for the values of \(x\). For the function \(y = xe^{1/x}\), solving \(y' = e^{1/x} - xe^{1/x}/x^2 = 0\) gives us critical points at \(x=1\) and \(x=-1\).

Analyzing the behavior of the function at these points helps determine whether they are local maxima, minima, or saddle points.
Product Rule
The product rule is a fundamental rule of calculus used to find the derivative of the product of two functions. If you have two differentiable functions, say \(u(x)\) and \(v(x)\), the derivative of their product \(u(x)v(x)\) is given by:
  • Find the derivative of each function separately.
  • Apply the product rule: \( (uv)' = u'v + uv' \).
In our example, \(y = xe^{1/x}\), the product rule gives us the first derivative: \(y' = e^{1/x} - xe^{1/x}/x^2\). This derivative is then used to find critical points and analyze the behavior of the function for local extrema.
Function Analysis
Function analysis involves studying a function to understand its characteristics and behavior. It includes finding extrema, analyzing intervals of increase and decrease, concavity, and other properties.

Using the first derivative, we can:
  • Determine intervals where the function increases or decreases.
  • Identify potential maxima and minima.
In our exercise with \(y = xe^{1/x}\), the function decreases on intervals \((-鈭�, -1)\) and \((-1, 1)\), then increases on \( (1, +鈭�)\). This behavior confirms \(x=-1\) as a local maximum and \(x=1\) as a local minimum. Understanding these concepts is essential for thorough function analysis.

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Most popular questions from this chapter

Tin Pest When metallic tin is kept below \(13.2^{\circ} \mathrm{C}\) it slowly becomes brittle and crumbles to a gray powder. Tin objects eventually crumble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious. And indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that controls the rate of reaction without undergoing any permanent change in itself. An autocatalytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at first if the amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abundant, the reaction proceeds at a faster pace. In some cases it is reasonable to assume that the rate \(v=d x / d t\) of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, \(v\) may be considered to be a function of \(x\) alone, and $$v=k x(a-x)=k a x-k x^{2}$$ where \(\begin{aligned} x &=\text { the amount of product, } \\ a &=\text { the amount of substance at the beginning, } \\ k &=\text { a positive constant. } \end{aligned}\) At what value of \(x\) does the rate \(v\) have a maximum? What is the maximum value of \(v ?\)

Draining Hemispherical Reservoir Water is flowing at the rate of 6 \(\mathrm{m}^{3} / \mathrm{min}\) from a reservoir shaped like a hemispherical bowl of radius \(13 \mathrm{m},\) shown here in profile. Answer the following questions given that the volume of water in a hemispherical bowl of radius \(R\) is \(V=(\pi / 3) y^{2}(3 R-y)\) when the water is \(y\) units deep. (a) At what rate is the water level changing when the water is 8 m deep? (b) What is the radius \(r\) of the water's surface when the water is \(y\) m deep? (c) At what rate is the radius \(r\) changing when the water is 8 \(\mathrm{m}\) deep?

Expanding Circle The radius of a circle is increased from 2.00 to 2.02 \(\mathrm{m} .\) (a) Estimate the resulting change in area. (b) Estimate as a percentage of the circle's original area.

$$ \begin{array}{l}{\text { Multiple Choice All of the following functions satisfy the }} \\ {\text { conditions of the Mean Value Theorem on the interval }[-1,1]} \\ {\text { except } }\end{array} $$ \((\mathbf{A}) \sin x\) \((\mathbf{B}) \sin ^{-1} x\) \((\mathrm{C}) x^{5 / 3}\) (D) \(x^{3 / 5}\) (E) \(\frac{x}{x-2}\)

Group Activity Cardiac Output In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurtzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 liters a minute. At rest it is likely to be a bit under 6 \(\mathrm{L} / \mathrm{min}\) . If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min.}\) Your cardiac output can be calculated with the formula $$$=\frac{Q}{D}$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{mL} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{mL} / \mathrm{min}\) and \(D=97-56=41 \mathrm{mL} / \mathrm{L}\) $$y=\frac{233 \mathrm{mL} / \mathrm{min}}{41 \mathrm{mL} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}$$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D. Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?
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