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Chapter 4: Problem 46

\(\begin{equation}\begin{array}{l}{\text { a. Identify the function's local extreme values in the given }} \\ {\text { domain, and say where they occur. }} \\ {\text { b. Which of the extreme values, if any, are absolute? }} \\\ {\text { c. Support your findings with a graphing calculator or computer }} \\\ {\text { grapher. }}\end{array}\end{equation}\) $$\begin{equation} f(t)=t^{3}-3 t^{2}, \quad-\infty < t \leq 3 \end{equation}$$

Short Answer

Expert verified
The local minimum is at \( t = 2 \) with a value of -4, which is also an absolute minimum. There is no absolute maximum in \( -\infty < t \leq 3 \).

Step by step solution

01

Find the critical points

To find the local extrema, we need to find the derivative of \( f(t) = t^3 - 3t^2 \). The derivative \( f'(t) = 3t^2 - 6t \). Set \( f'(t) = 0 \) to find critical points: \( 3t(t - 2) = 0 \). Solving this gives \( t = 0 \) and \( t = 2 \) as critical points.
02

Evaluate the function at critical points

Substitute the critical points into the function to find their values: \( f(0) = 0^3 - 3(0)^2 = 0 \) and \( f(2) = 2^3 - 3(2)^2 = 8 - 12 = -4 \). Hence, the critical points are (0, 0) and (2, -4).
03

Evaluate the function at the boundary

Since \( t \leq 3 \), evaluate the function at \( t = 3 \). \( f(3) = 3^3 - 3(3)^2 = 27 - 27 = 0 \). Thus, at the boundary, the value is \( f(3) = 0 \).
04

Determine the nature of the critical points

To classify each critical point, consider the interval sign changes: for \( t = 2 \), test intervals \((-\infty, 0)\), \((0, 2)\), \((2, 3)\). For \( t = 2 \), \( f'(t) \) changes from negative to positive, indicating a local minimum at (2, -4). Since \( t=0,3 \) provide the same value \( f(0) = f(3) = 0 \), these are neither absolute maxima nor minima in this domain.
05

Determine absolute extreme values

The function achieves its lowest value at \( t = 2 \) with \( f(2) = -4 \), which is the absolute minimum. At \( t = 0 \) and \( t = 3 \), it reaches 0, which is not the lowest since another point is lower; thus, neither of these is an absolute maximum.
06

Graphing support

Using a graphing tool, plot \( f(t) = t^3 - 3t^2 \). You will observe a curve with a trough at \( t = 2 \). The graph shows the lowest point at \( t = 2, f(t) = -4 \), confirming it is the absolute minimum, while neither 0 nor 3 provides higher values than the entire increasing right of the minimum.

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Key Concepts

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

Local Extrema
A critical point of a function is where its derivative is zero or undefined, indicating a potential local maximum or minimum. To determine local extrema, we first find the derivative of the function. In our case, let's consider the function \( f(t) = t^3 - 3t^2 \). The derivative is \( f'(t) = 3t^2 - 6t \). We set this equal to zero to find critical points: \( 3t(t-2) = 0 \), giving us \( t = 0 \) and \( t = 2 \) as critical points.

At each critical point, substitute back into the original function to find the function's value at these points:
  • At \( t = 0 \), \( f(0) = 0 \).
  • At \( t = 2 \), \( f(2) = -4 \).
Finally, evaluate at the boundary of the domain, \( t = 3 \), which gives \( f(3) = 0 \).

Examining these points, \( t = 2 \) emerges as a local minimum, while \( t = 0 \) and \( t = 3 \) both correspond to a value of zero, not indicating maxima or minima.
Derivative
Understanding derivatives is crucial when analyzing critical points and local extrema. The derivative of a function provides us with a formula to find how the function changes at any point along its curve. It is a powerful tool to find where these changes cease, like at peaks, valleys, or points of inflection in the graph.

For \( f(t) = t^3 - 3t^2 \), the derivative \( f'(t) = 3t^2 - 6t \) gives us the slope of the tangent to the function at any point \( t \).

Calculating where \( f'(t) = 0 \) will pinpoint where potential maxima, minima, or inflections occur:
  • The critical points are calculated as \( t = 0 \) and \( t = 2 \).
These points suggest where the function changes direction, revealing crucial insights into the nature of the function. Hence, understanding and using derivatives allows us to identify significant characteristics of the graph and solve real-world optimization problems.
Graphical Analysis
Graphical analysis plays an integral role in understanding the behavior of functions in calculus. By visually examining \( f(t) = t^3 - 3t^2 \), we see the curve's dynamics, including peaks and troughs. Use a graphing tool or calculator to plot the function and observe how it bends and turns in relation to the x-axis.

For our function, the graph illustrates the critical points at \( t = 0 \), \( t = 2 \), and around the boundary \( t = 3 \):
  • The graph dips sharply at \( t = 2 \), confirming a local minimum at \( -4 \).
  • The values at \( t = 0 \) and \( t = 3 \) are both \( 0 \), where the function returns to the x-axis.
Graphically identifying these points helps validate our analytical findings, reinforcing the calculus outcomes with visual proof. This dual approach of algebraic and graphical analysis ensures a thorough understanding of the function's behavior across its domain.

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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}

Identify the inflection points and local maxima and minima of the functions graphed. Identify the intervals on which the functions are concave up and concave down. $$y=\frac{9}{14} x^{1 / 3}\left(x^{2}-7\right)$$

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 \(v ?\)

The hammer and the feather When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about 4 ft above the ground. The television footage of the event shows the hammer and the feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the 4 ft. How long did it take the hammer and feather to fall 4 ft on the moon? To find out, solve the following initial value problem for \(s\) as a function of \(t\) . Then find the value of \(t\) that makes \(s\) equal to 0 . \begin{equation} \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} s}{d t^{2}}=-5.2 \mathrm{ft} / \mathrm{sec}^{2}} \\ {\text { Initial conditions: }} & {\frac{d s}{d t}=0 \text { and } s=4 \text { when } t=0}\end{array} \end{equation}

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?
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