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Chapter 4: Problem 22

\(\begin{equation} \begin{array}{l}{\text { a. Find the open intervals on which the function is increasing and }} \\ {\text { decreasing. }} \\ {\text { b. Identify the function's local and absolute extreme values, if }} \\ {\text { any, saying where they occur. }}\end{array} \end{equation}\) $$\begin{equation} h(x)=2 x^{3}-18 x \end{equation}$$

Short Answer

Expert verified
The function is increasing on \(( -\infty, -\sqrt{3})\) and \(( \sqrt{3}, \infty)\); decreasing on \((-\sqrt{3}, \sqrt{3})\). Local max at \( (-\sqrt{3}, 12\sqrt{3}) \), local min at \( (\sqrt{3}, -12\sqrt{3}) \). No absolute extremes.

Step by step solution

01

Find the First Derivative

To determine the intervals where the function is increasing or decreasing, we need to find the first derivative of the function. The function given is \( h(x) = 2x^3 - 18x \). Using the power rule, the derivative is \( h'(x) = 6x^2 - 18 \).
02

Find Critical Points

To find the critical points, set the first derivative \( h'(x) = 6x^2 - 18 \) to zero and solve for \( x \): \( 6x^2 - 18 = 0 \). Factor to get \( 6(x^2 - 3) = 0 \), thus \( x^2 = 3 \). Solving gives critical points at \( x = \sqrt{3} \) and \( x = -\sqrt{3} \).
03

Determine Intervals of Increase and Decrease

Use the critical points to divide the real line into intervals: \(( -\infty, -\sqrt{3} )\), \(( -\sqrt{3}, \sqrt{3} )\), and \(( \sqrt{3}, \infty )\). Test each interval using a point within it in the first derivative \( h'(x) \).For \( x = -2 \) in \((-\infty, -\sqrt{3})\), \( h'(-2) = 6(-2)^2 - 18 = 6 \times 4 - 18 = 6 > 0 \), increasing.For \( x = 0 \) in \((-\sqrt{3}, \sqrt{3})\), \( h'(0) = -18 < 0 \), decreasing.For \( x = 2 \) in \((\sqrt{3}, \infty)\), \( h'(2) = 6 \times 4 - 18 = 6 > 0 \), increasing.
04

Identify Local Extreme Values

To find local extreme values, examine the behavior of \( h(x) \) around the critical points. A local maximum or minimum occurs where the derivative changes sign.At \( x = -\sqrt{3} \), \( h(x) \) changes from increasing to decreasing, so there is a local maximum.At \( x = \sqrt{3} \), \( h(x) \) changes from decreasing to increasing, so there is a local minimum.Calculate the corresponding \( h(x) \) values:\[ h(-\sqrt{3}) = 2(-\sqrt{3})^3 - 18(-\sqrt{3}) = -6\sqrt{3} + 18\sqrt{3} = 12\sqrt{3} \] \[ h(\sqrt{3}) = 2(\sqrt{3})^3 - 18(\sqrt{3}) = 6\sqrt{3} - 18\sqrt{3} = -12\sqrt{3} \]
05

Determine Absolute Extreme Values

Because the function \( h(x) = 2x^3 - 18x \) is a polynomial of degree 3, it extends to infinity in both directions, so there are no absolute maximum or minimum values on the entire real line.

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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 of a function, often written as \( f'(x) \), is a powerful tool in calculus to understand how a function behaves. It tells us about the rate at which the function's value is changing. If \( f'(x) > 0 \), the function is increasing at that point. If \( f'(x) < 0 \), the function is decreasing. Let's break it down further:
  • Power Rule: To find the derivative of \( h(x)=2x^3-18x \), we apply the power rule. The power rule says that if you have a term like \( ax^n \), the derivative is \( nax^{n-1} \).
  • Calculation: Using the power rule, the derivative of \( 2x^3 \) is \( 6x^2 \) and \( 18x \) is \( 18 \). So, the first derivative is \( h'(x)=6x^2-18 \).
  • Interpreting First Derivative: We use this derivative to find where the function increases or decreases by checking the sign of this derivative over certain intervals.
Critical Points
Critical points are specific values of \( x \) where the first derivative is equal to zero or undefined. These points are crucial because they can indicate where the function might have maxima, minima, or saddle points.
  • Finding Critical Points: To find critical points, we solve \( h'(x) = 6x^2 - 18 = 0 \). This simplifies to \( x^2 = 3 \), giving us \( x = \sqrt{3} \) and \( x = -\sqrt{3} \).
  • Why Important: These points mark where the function's slope (or rate of change) transitions, and they are essential for analyzing the function's behavior, especially when looking for local maximums or minimums.
Once the critical points are found, they can be used to determine where the function is increasing or decreasing by testing intervals around these points.
Local Extrema
Local extrema refer to the local maximum and minimum values of a function, which occur at critical points where the derivative changes sign.
  • Definition: A local maximum is a point where the function value is higher than those of its neighboring points, and a local minimum is where it is lower.
  • Changes in Derivative: For a local maximum at \( x = -\sqrt{3} \), \( h(x) \) transitions from increasing to decreasing. For a local minimum at \( x = \sqrt{3} \), it changes from decreasing to increasing.
  • Value Calculation: At \( x = -\sqrt{3} \), \( h(x) = 12\sqrt{3} \) indicates a local maximum. At \( x = \sqrt{3} \), \( h(x) = -12\sqrt{3} \) suggests a local minimum.
By identifying these points and checking the derivative signs, we can understand the function's peaks and valleys better. This analysis is crucial for graphing functions and predicting behavior.

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Most popular questions from this chapter

Stopping a car in time You are driving along a highway at a steady 60 \(\mathrm{mph}(88 \mathrm{ft} / \mathrm{sec})\) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in 242 \(\mathrm{ft}\) ? To find out, carry out the following steps. \begin{equation} \begin{array}{l}{\text { 1. Solve the initial value problem }} \\ {\text { Differential equation: } \frac{d^{2} s}{d t^{2}}=-k \quad(k \text { constant })} \\ {\text { Initial conditions: } \frac{d s}{d t}=88 \text { and } s=0 \text { when } t=0}\\\\{\text { 2. Find the value of } t \text { that makes } d s / d t=0 . \text { (The answer will }} \\ {\text { involve } k . )} \\\ {\text { 3. Find the value of } k \text { that makes } s=242 \text { for the value of } t \text { you }} \\ {\text { found in Step } 2 \text { . }}\end{array} \end{equation}

Use a CAS to solve the initial value problems in Exercises \(107-110\) . Plot the solution curves. $$y^{\prime}=\cos ^{2} x+\sin x, \quad y(\pi)=1$$

Finding displacement from an antiderivative of velocity a. Suppose that the velocity of a body moving along the \(s\) -axis is \begin{equation} \begin{array}{c}{\text { a. Suppose that the velocity of a body moving along the } s \text { -axis is }} \\ {\frac{d s}{d t}=v=9.8 t-3} \\ {\text { i) Find the body's displacement over the time interval from }} \\ {t=1 \text { to } t=3 \text { given that } s=5 \text { when } t=0 \text { . }}\\\\{\text { ii) Find the body's displacement from } t=1 \text { to } t=3 \text { given }} \\\ {\text { that } s=-2 \text { when } t=0 \text { . }} \\ {\text { iii) Now find the body's displacement from } t=1 \text { to } t=3} \\ {\text { given that } s=s_{0} \text { when } t=0 \text { . }}\\\\{\text { b. Suppose that the position } s \text { of a body moving along a coordinate }} \\ {\text { line is a differentiable function of time } t . \text { Is it true that }} \\\ {\text { once you know an antiderivative of the velocity function }}\\\\{d s / d t \text { you can find the body's displacement from } t=a \text { to }} \\\ {t=b \text { even if you do not know the body's exact position at }} \\\ {\text { either of those times? Give reasons for your answer. }}\end{array} \end{equation}

Let \(f\) be differentiable at every value of \(x\) and suppose that \(f(1)=1,\) that \(f^{\prime} < 0\) on \((-\infty, 1),\) and that \(f^{\prime} > 0\) on \((1, \infty)\) . a. Show that \(f(x) \geq 1\) for all \(x\) b. Must \(f^{\prime}(1)=0 ?\) Explain.

Find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function's first and second derivatives. How are the values at which these graphs intersect the \(x\)-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? \(y=x^{5}-5 x^{4}-240\)
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