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Chapter 4: Problem 36

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x^{2 / 3}(x-5)$$

Short Answer

Expert verified
Local minimum at \( x = 2 \); inflection point at \( x = -1 \).

Step by step solution

01

Find the First Derivative

To find extremes, calculate the first derivative of the function. The given function is \( y = x^{2/3}(x-5) \). Use the product rule: \( u = x^{2/3} \) and \( v = x - 5 \). The derivatives: \( u' = \frac{2}{3}x^{-1/3} \) and \( v' = 1 \). So, \( y' = u'v + uv' = \frac{2}{3}x^{-1/3}(x-5) + x^{2/3} \cdot 1 \). This simplifies to: \( y' = \frac{2}{3}x^{2/3} - \frac{10}{3}x^{-1/3} + x^{2/3} \), ultimately \( y' = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} \).
02

Identify Critical Points

Critical points occur where \( y' = 0 \) or \( y' \) is undefined. Set \( \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = 0 \). Factor out \( x^{-1/3} \): \( x^{-1/3}(\frac{5}{3}x - \frac{10}{3}) = 0 \). Thus, \( x = 0 \) or \( x = 2 \). Check for undefined points; none in the domain \( x > 0 \).
03

Determine Local Extremes from Critical Points

Use the first derivative test. Test intervals around critical points \( x = 0 \) and \( x = 2 \). For \( x < 0 \), choose \( x = -1 \): \( y'(-1) < 0 \). For \( 0 < x < 2 \), choose \( x = 1 \): \( y'(1) < 0 \). For \( x > 2 \), choose \( x = 3 \): \( y'(3) > 0 \). So, a local minimum at \( x = 2 \).
04

Find the Second Derivative for Inflection Points

Calculate the second derivative. From \( y' = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} \), differentiate again. \( y'' = \frac{10}{9}x^{-1/3} + \frac{10}{9}x^{-4/3} \). Set \( y'' = 0 \) to find inflection points. Factor \( \frac{10}{9}x^{-4/3}(x + 1) = 0 \). Hence, \( x = -1 \).
05

Verify Inflection Points

Check the sign change of \( y'' \) around \( x = -1 \). For \( x < -1, \) choose \( x = -2 \): \( y''(-2) > 0. \) For \( x > -1, \) choose \( x = 0 \): \( y''(0) < 0 \). Since \( y'' \) changes sign, \( x = -1 \) is an inflection point.
06

Graph the Function

Plot the function considering the identified points: the local minimum at \( x = 2 \), inflection point at \( x = -1 \), and behavior as \( x \to \infty \) and \( x \to 0 \). Use these to sketch the curve.

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Key Concepts

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

Critical Points
Critical points are where a function's derivative is zero or undefined. These points might be where a curve changes direction, defining peaks or troughs. For the function given, critical points are found by setting the first derivative to zero and solving:
  • Start with the derivative: \( y' = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} \).
  • Set \( y' = 0 \) to find points: \( x^{-1/3}(\frac{5}{3}x - \frac{10}{3}) = 0 \).
  • This leads to \( x = 0 \) or \( x = 2 \).
By solving the equation, we found the critical points at \( x = 0 \) and \( x = 2 \). These points are examined further to assess if they're extremes.
Local Extremes
Local extremes refer to local maximums or minimums of a function. At these points, the slope of the tangent to the function's graph is zero. Once critical points are identified, use the first derivative test to determine whether each is a maximum or minimum:
  • Evaluate the sign of \( y' \) around critical points.
  • For \( x = 0 \), test values show \( y' < 0 \) around this point.
  • For \( x = 2 \), test values change from negative \( y' \) before \( x = 2 \) to positive after, indicating a local minimum.
This analysis indicates that \( x = 2 \) is a local minimum, where the function's curve dips downlowest locally.
Inflection Points
Inflection points are where a function's graph changes its concavity. This means the curve goes from bending upwards to downwards, or vice versa. These points are found using the second derivative:
  • Calculate the second derivative: \( y'' = \frac{10}{9}x^{-1/3} + \frac{10}{9}x^{-4/3} \).
  • Set \( y'' = 0 \) to find inflection points: factor \( \frac{10}{9}x^{-4/3}(x + 1) = 0 \).
  • Solution gives \( x = -1 \).
Check around \( x = -1 \) to assure a sign change in \( y'' \). Before, \( y'' > 0 \) and after, \( y'' < 0 \), confirming \( x = -1 \) is indeed an inflection point.
Derivative
The derivative tells us how a function changes as you move along its graph. For our function, taking the derivative helps find rates of change and critical points. Here's a quick summary on how derivatives work:
  • The first derivative, \( y' \), indicates slope and helps find critical points for potential extremes.
  • The second derivative, \( y'' \), gives insight into concavity, revealing inflection points.
In short, derivatives provide a powerful way to analyze a function's behavior by looking at how its rate of change itself changes.
Graphing Functions
Graphing a function involves visually representing its mathematical behavior. Here's how the key calculus concepts from the earlier steps assist in sketching the function's curve:
  • Identify key coordinates like local minimums and inflection points to mark on the graph.
  • The critical points \( x = 0 \) and \( x = 2 \) guide where the curve bends or twists.
  • Inflection point at \( x = -1 \) shows where the curve changes direction in concavity.
Once these points are plotted, the graph should accurately reflect the increasing or decreasing nature and shifts in concavity of the function. The graph considers key behaviors like what happens as \( x \to \infty \) and its behavior near zero.

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Most popular questions from this chapter

Motion along a coordinate line \(A\) particle moves on a coordinate line with acceleration \(a=d^{2} s / d t^{2}=15 \sqrt{t}-(3 / \sqrt{t})\) subject to the conditions that \(d s / d t=4\) and \(s=0\) when \(t=1\) Find \begin{equation} \begin{array}{l}{\text { a. the velocity } v=d s / d t \text { in terms of } t} \\ {\text { b. the position } s \text { in terms of } t \text { . }}\end{array} \end{equation}

Suppose the derivative of the function \(y=f(x)\) is $$y^{\prime}=(x-1)^{2}(x-2)(x-4).$$ At what points, if any, does the graph of \(f\) have a local minimum, local maximum, or point of inflection?

Let \(f\) be a function defined on an interval \([a, b] .\) What conditions could you place on \(f\) to guarantee that $$\min f^{\prime} \leq \frac{f(b)-f(a)}{b-a} \leq \max f^{\prime}$$ where min \(f^{\prime}\) and max \(f^{\prime}\) refer to the minimum and maximum values of \(f^{\prime}\) on \([a, b] ?\) Give reasons for your answers.

Suppose that the second derivative of the function \(y=f(x)\) is $$y^{\prime \prime}=x^{2}(x-2)^{3}(x+3).$$ For what \(x\)-values does the graph of \(f\) have an inflection point?

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 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 \(v ?\)
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