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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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