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Chapter 4: Problem 35

In Exercises \(35-42,\) identify the critical point and determine the local extreme values. $$y=x^{2 / 3}(x+2)$$

Short Answer

Expert verified
The local maximum values are at \(x = -2\) with \(y = 0\) and at \(x = 2/5\) with \(y = 312/125\). The local minimum value is at \(x = 0\) with \(y = 0\).

Step by step solution

01

Differentiate the function

Find the derivative of \(y = x^{2/3}(x+2)\) using the power rule, product rule, and chain rule if necessary. The differentiation will give \(y' = \frac{5}{3}x^{2/3} - \frac{4}{3}x^{-1/3}\).
02

Find the critical points

Set the derivative equal to zero and solve for \(x\). The solutions to the equation \(\frac{5}{3}x^{2/3} - \frac{4}{3}x^{-1/3} = 0\) are \(x = -2\), \(x = 0\), and \(x = 2/5\). These values of \(x\) are the critical points of the function.
03

Determine the local extreme values

Using the first derivative test, plug in values into the derivative that are less than and greater than the critical points. From the sign of the derivative at those points, finding whether the function is increasing or decreasing around the critical points will determine if the critical points are local maxima, local minima, or neither. After testing, it is discovered that the function increases up to \(x = -2\), then decreases until \(x = 0\), then increases until \(x = 2/5\), then decreases. Therefore the local maximum occurs at \(x = -2\) and \(x =2/5\) and the local minimum occurs at \(x = 0\).

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Key Concepts

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

Critical Points
In calculus, finding critical points is an essential step in analyzing the behavior of a function. Critical points are values of \(x\) where the derivative of the function is zero or undefined. These points are crucial as they can indicate potential maxima or minima for the function.
To find critical points, one must:
  • Take the derivative of the function.
  • Set the derivative equal to zero and solve for \(x\).
In some cases, also check where the derivative is undefined, as this still indicates a critical point. In our exercise, solving the equation \(\frac{5}{3}x^{2/3} - \frac{4}{3}x^{-1/3} = 0\) gave us the critical points \(x = -2\), \(x = 0\), and \(x = 2/5\). These points are where significant changes in behavior may occur.
Local Extrema
Local extrema refer to points in a function where the function takes either a local maximum or minimum value. They are closely associated with critical points. A local maximum is a point where the function value is greater than all nearby points, while a local minimum is where it is less than all nearby points.
To determine local extrema, analyze the function's derivative around its critical points. In our function, it was found that:
  • The local maximum occurs at \(x = -2\) and \(x = 2/5\).
  • The local minimum occurs at \(x = 0\).
These extrema tell us where the function's graph peaks or dips, which is vital for understanding its overall shape.
Derivative
The derivative of a function represents its rate of change or the slope of the tangent line at any given point. It's a fundamental concept in calculus as it helps determine the behavior of functions.
To differentiate \(y = x^{2/3}(x+2)\), we use differentiation rules:
  • The product rule, as the function is a product of two terms: \(x^{2/3}\) and \(x+2\).
  • The power rule for differentiating terms like \(x^{2/3}\) and \(x^{-1/3}\).
The resulting derivative \(y' = \frac{5}{3}x^{2/3} - \frac{4}{3}x^{-1/3}\) provides us with the necessary insights to locate critical points and perform further analysis.
First Derivative Test
The first derivative test is a method used to determine whether a critical point is a local maximum, minimum, or neither. It involves evaluating the sign of the derivative before and after the critical points. Here's how it's done:
1. Identify the critical points where the derivative is zero or undefined.
2. Test the sign of the derivative in intervals around each critical point.
  • If the derivative changes from positive to negative, it's a local maximum.
  • If it changes from negative to positive, it's a local minimum.
In the exercise, this test showed that our function increases up to \(x = -2\), decreases until \(x = 0\), then increases again until \(x = 2/5\), and finally decreases. This pattern helped identify the local extrema at those points, giving a clearer picture of the function's behavior.

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Most popular questions from this chapter

How We Cough When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the question of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v(\) in \(\mathrm{cm} / \mathrm{sec})\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in \(\mathrm{cm}\) and \(c\) is a positive constant whose value depends in part on the length of the trachea. (a) Show that \(v\) is greatest when \(r=(2 / 3) r_{0},\) that is, when the trachea is about 33\(\%\) contracted. The remarkable fact is that \(X\) -ray photographs confirm that the trachea contracts about this much during a cough. (b) Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see to the claim that \(v\) is a maximum when \(r=(2 / 3) r_{0}\) .

Multiple Choice If Newton's method is used to find the zero of \(f(x)=x-x^{3}+2,\) what is the third estimate if the first estimate is 1\(?\) \((\mathbf{A})-\frac{3}{4} \quad(\mathbf{B}) \frac{3}{2} \quad(\mathbf{C}) \frac{8}{5} \quad(\mathbf{D}) \frac{18}{11}\) \((\mathbf{E}) 3\)

Boring a Cylinder The mechanics at Lincoln Automotive are reboring a 6 -in. deep cylinder to fit a new piston. The machine they are using increases the cylinder's radius one-thousandth of an inch every 3 min. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.800 in.?

Writing to Learn Explain why there is a zero of \(y=\cos x\) between every two zeros of \(y=\sin x .\)

Measuring Acceleration of Gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\) . The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g\) . By keeping track of \(\Delta T\) , we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L .\) (a) With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts \((b)\) and \((c)\) . (b) Writing to Learn If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. (c) A clock with a 100 -cm pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001 \mathrm{sec} .\) Find \(d g\) and estimate the value of \(g\) at the new location.
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