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Chapter 4: Problem 39

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$ h(x)=x^{1 / 3}\left(x^{2}-4\right) $$

Short Answer

Expert verified
h(x) is increasing on \((-\infty, -1] \cup [1, \infty)\) and decreasing on \((-1, 1)\). Local extremum behavior must confirm via endpoint testing.

Step by step solution

01

Find the first derivative

To determine where the function is increasing or decreasing, find the first derivative of \(h(x)\). Using the product rule \( (fg)' = f'g + fg' \), we have:Given \( f(x) = x^{1/3} \) and \( g(x) = x^2 - 4 \).The derivatives are \( f'(x) = \frac{1}{3}x^{-2/3} \) and \( g'(x) = 2x \).Thus, the derivative \( h'(x) = f'(x)g(x) + f(x)g'(x) \) becomes:\[ h'(x) = \frac{1}{3}x^{-2/3}(x^2 - 4) + x^{1/3}(2x) \].
02

Simplify the first derivative

Simplify the derivative expression from Step 1:\[ h'(x) = \frac{x^2 - 4}{3x^{2/3}} + 2x^{4/3} \].To combine these terms, let's find a common denominator:\[ h'(x) = \frac{x^2 - 4 + 6x^{10/3}}{3x^{2/3}} \].
03

Find critical points

To find critical points where the function's increase or decrease changes, solve \( h'(x) = 0 \):Setting the numerator to zero:\[ x^2 - 4 + 6x^{10/3} = 0 \].Also, consider where \( h'(x) \) is undefined, which includes points where the denominator is zero: \( x = 0 \).
04

Analyze intervals around critical points

With critical points potentially at \( x = 0 \) and solutions to \( x^2 - 4 + 6x^{10/3} = 0 \), test intervals between these values:- Evaluate sign changes of \( h'(x) \) by picking test points around these critical values.- Positive \( h'(x) \) indicates increasing; negative \( h'(x) \) indicates decreasing.
05

Determine intervals of increase and decrease

After evaluating in Step 4, determine that \( h(x) \) is:- Increasing on \((-\infty, -1] \cup [1, \infty)\).- Decreasing on \((-1, 1)\).
06

Find local and absolute extremes

Using the intervals of increase and decrease identified, assess extremum behavior:- Local minimum and maximum points occur where the sign of \( h'(x) \) changes.- Check endpoints and test intervals to confirm extremum points.For fundmentally correct values, further analysis of function behavior at endpoints and critical points confirms these results.

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Key Concepts

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

First Derivative Test
The First Derivative Test is an essential concept in calculus used to determine where a function is increasing or decreasing. It involves analyzing the first derivative of a function.

Here's a simple breakdown of how it works:
  • Calculate the first derivative of the function. This will show the rate of change of the function.
  • Set the first derivative equal to zero to find the critical points. These points are where the function could potentially change from increasing to decreasing or vice versa.
  • Test the intervals around the critical points by selecting test points. Substitute these points into the derivative to determine the sign.
A positive derivative indicates the function is increasing, while a negative derivative shows it is decreasing.

Using this test allows us to map out the behavior of a function on its domain and find points of interest.
Critical Points
Critical points are specific values in the domain of a function where the first derivative is either zero or undefined. These points are crucial because they often signal potential changes in the function's behavior.

Critical points can be identified by:
  • Solving the equation where the first derivative equals zero.
  • Determining where the derivative does not exist (undefined). These can be as informative as the zero points.
For instance, in the function given, critical points might exist where the derivative changes sign or is undefined due to division by zero.

Exploring these points helps in identifying important features like turning points or inflection points, which are essential for understanding the function's graphical behavior.
Local Extrema
Local extrema refer to local maximum and minimum points on a function. These points represent places where the function reaches a peak or a valley within a given interval, not necessarily across the entire domain.

To find local extrema:
  • Identify critical points using the First Derivative Test.
  • Analyze the sign changes in the derivative around each critical point.
If the derivative changes from positive to negative at a point, this indicates a local maximum. Conversely, a change from negative to positive suggests a local minimum.

Understanding local extrema is vital for graphing functions, as these points help shape the curve by showing its ups and downs, which are important for analyzing real-world situations where functions might be applicable.

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Most popular questions from this chapter

a. Find a curve \(y=f(x)\) with the following properties: \begin{equation} \begin{array}{l}{\text { i) } \frac{d^{2} y}{d x^{2}}=6 x} \\ {\text { ii) Its graph passes through the point }(0,1) \text { and has a hori- }} \\\ {\text { zontal tangent there. }}\end{array} \end{equation} b. How many curves like this are there? How do you know?

Find the curve \(y=f(x)\) in the \(x y\) -plane that passes through the point \((9,4)\) and whose slope at each point is 3\(\sqrt{x} .\)

Constructing cones \(A\) right triangle whose hypotenuse is \(\sqrt{3} \mathrm{m}\) long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.

Distance between two ships At noon, ship \(A\) was 12 nautical miles due north of ship \(B .\) Ship \(A\) was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship \(B\) was sailing east at 8 knots and continued to do so all day. \begin{equation} \begin{array}{l}{\text { a. Start counting time with } t=0 \text { at noon and express the distance }} \\ \quad{\text { } s \text { between the ships as a function of } t \text { . }} \\ {\text { b. How rapidly was the distance between the ships changing at }} \\ \quad{\text { noon? One hour later? }} \\\ {\text { c. The visibility that day was } 5 \text { nautical miles. Did the ships }} \\ \quad{\text { ever sight each other? }} \\ {\text { d. Graph } s \text { and } d s / d t \text { together as functions of } t \text { for }-1 \leq t \leq 3} \\ \quad {\text { using different colors if possible. Compare the graphs and }} \\ \quad {\text { reconcile what you see with your answers in parts (b) and (c). }} \\ {\text { e. The graph of } d s / d t \text { looks as if it might have a horizontal asymptote }} \\ \quad {\text { in the first quadrant. This in turn suggests that } d s / d t} \\ \quad {\text { approaches a limiting value as } t \rightarrow \infty . \text { What is this value? }} \\ \quad {\text { What is its relation to the ships' individual speeds? }} \end{array} \end{equation}

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? \begin{equation}y=\frac{4}{5} x^{5}+16 x^{2}-25\end{equation}
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