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In Exercises \(37-40,\) find the function's absolute maximum and minimum values and say where they are assumed. $$ g(\theta)=\theta^{3 / 5}, \quad-32 \leq \theta \leq 1 $$

Short Answer

Expert verified
Absolute maximum is 1 at \( \theta = 1 \); absolute minimum is -8 at \( \theta = -32 \).

Step by step solution

01

Identify Critical Points

First, we need to find the derivative of the function \( g(\theta) = \theta^{3/5} \). The derivative is given by: \( g'(\theta) = \frac{3}{5} \theta^{-2/5} \). The critical points occur where the derivative is zero or undefined. Here, the derivative is undefined at \( \theta = 0 \) because the power of the term becomes negative.
02

Evaluate the Function at Critical Points and Endpoints

The critical point is \( \theta = 0 \), and the endpoints of the interval are \( \theta = -32 \) and \( \theta = 1 \). Now, evaluate \( g(\theta) \) at these points: - \( g(-32) = (-32)^{3/5} = -8 \) (since \((-32)^{1/5} = -2\))- \( g(0) = 0^{3/5} = 0 \)- \( g(1) = 1^{3/5} = 1 \).
03

Determine Absolute Maximum and Minimum

Compare the function values calculated at the endpoints and critical points: - \( g(-32) = -8 \)- \( g(0) = 0 \)- \( g(1) = 1 \)The absolute maximum value is 1 at \( \theta = 1 \) and the absolute minimum value is -8 at \( \theta = -32 \).

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Key Concepts

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

Understanding Critical Points
Critical points are essential for determining where a function might reach its highest or lowest values within an interval. These points occur where the derivative of the function is either zero or undefined. A derivative equals zero implies a flat tangent line, suggesting a potential maxima or minima. When a derivative is undefined, it might indicate a cusp or vertical tangent.
In the example of the function \( g(\theta) = \theta^{3/5} \), the critical point is found not by setting the derivative equal to zero but by identifying where it becomes undefined. The derivative \( g'(\theta) = \frac{3}{5} \theta^{-2/5} \) is undefined at \( \theta = 0 \).
This becomes a critical point because the exponent causes division by zero in the derivative calculation. Recognizing these cases can greatly aid in understanding the behavior of functions.
The Role of the Derivative
In calculus, the derivative of a function is a powerful tool that provides insights about the function's behavior. It signifies the rate at which the function value changes as the variable progresses. When determining absolute extrema, the derivative serves to identify critical points.
For \( g(\theta) = \theta^{3/5} \), the derivative is calculated as \( g'(\theta) = \frac{3}{5} \theta^{-2/5} \). This derivative helps pinpoint where the changes in slope—potential maximums or minimums—occur and where the slope could be undefined.
Calculating a derivative involves applying the rules of differentiation, such as the power rule. Understanding these rules and their application is crucial for finding and interpreting critical points in any function.
Exploring Interval Endpoints
In addition to critical points, interval endpoints are focal spots for examining potential absolute extrema. When a function is defined on a closed interval, extrema can also occur right at the edges or boundaries of this interval.
In the case of \( g(\theta) = \theta^{3/5} \), the interval is from \(-32\) to \(1\). We must evaluate the function at both of these endpoints because they could represent the smallest or largest value within the interval.
Don't overlook endpoints during your evaluations. Including them ensures a comprehensive analysis of where a function might reach its extremum, complementing the assessments at critical points.
Carrying Out Function Evaluation
Function evaluation is the process of calculating the actual values of a function at certain specified points, such as critical points and endpoints. This step is essential for comparing these values to locate the absolute maximum and minimum.
In our example, evaluations of \( g(\theta) = \theta^{3/5} \) are performed at the critical point and endpoints, giving \( g(-32) = -8 \), \( g(0) = 0 \), and \( g(1) = 1 \). Comparing these values reveals that the absolute maximum is 1 at \( \theta = 1 \), and the minimum is -8 at \( \theta = -32 \).
Function evaluation grants us concrete data to analyze, completing the process of identifying extrema. It's a vital final step in confirming where and what the extrema are for a given interval.

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Most popular questions from this chapter

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?

Identify the inflection points and local maxima and minima of the functions graphed in Exercises \(1-8 .\) Identify the intervals on which the functions are concave up and concave down. $$ y=\frac{x^{3}}{3}-\frac{x^{2}}{2}-2 x+\frac{1}{3} $$

Quadratic curves What can you say about the inflection points of a quadratic curve \(y=a x^{2}+b x+c, a \neq 0 ?\) Give reasons for your answer.

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

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=4 x^{3}-x^{4}=x^{3}(4-x)$$

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