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Chapter 3: Problem 32

Locate the absolute extrema of the function (if any exist) over each interval. \(f(x)=\sqrt{4-x^{2}}\) (a) [-2,2] (b) [-2,0) (c) (-2,2) (d) [1,2)

Short Answer

Expert verified
The absolute extrema for the given intervals are: (a) [-2, 2] maximum at f(0) = 2, minimum at f(-2) = f(2) = 0; (b) [-2, 0) maximum at f(0) = 2, minimum at f(-2) = 0; (c) (-2, 2) the maximum at f(0) = 2, no minimum; and (d) [1, 2) maximum at f(1) = \(\sqrt{3}\), minimum at f(2) = 0.

Step by step solution

01

Finding the first derivative

To find where the absolute extrema occurs, it's necessary to find where the derivative equals 0 or is undefined. The derivative of \(f(x)=\sqrt{4-x^{2}}\), using the chain rule, is \(f'(x) = -x/\sqrt{4-x^{2}}\).
02

Finding the critical points

Set the derivative to 0 and solve for x, it's impossible because the fraction is 0 when the numerator is 0, but if \(x = 0\), the denominator is non-zero. Therefore, there's no real solutions, which means there's no critical points except the end points.
03

Substituting end points and critical points into original function

Substitute the endpoints of the interval and the critical points (if exist) into the original function. For the interval [-2,2], the endpoints are x = -2, 2. Therefore, the maximum and minimum values are the maximum and minimum of f(-2), f(2), which is also the maximum and minimum of -\(\sqrt{4}\), \(\sqrt{4}\). Repeat this step for each of the intervals [(a), (b), (c), and (d)].
04

Identifying the absolute extrema

The absolute extrema are the maximum and minimum values calculated in Step 3. For each interval, identify the smallest and largest values.

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Key Concepts

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

Absolute Extrema
When we talk about absolute extrema, we are referring to the highest or lowest points the function achieves on a given interval. These points are found by plugging in specific x-values into the function. Absolute extrema are crucial because they help us understand the behavior of a function over a particular range.
  • Absolute Maximum: Largest y-value on the interval
  • Absolute Minimum: Smallest y-value on the interval
We find absolute extrema by evaluating both the endpoints of an interval and the critical points of the function. This is important because a function could achieve its highest or lowest value at the bounds or within the interval. Hence, including all possible cases ensures accuracy in finding the extrema.
Critical Points
Critical points are the x-values where a function's derivative equals zero or is undefined. These are crucial in determining a function's behavior because they are potential points where the function's direction changes, which may correspond to a local maximum or minimum.
  • Set the derivative equal to zero and solve for x
  • If the derivative does not exist at a point, that point is also a critical point
It's essential to check these points as they could reveal places where the function significantly changes its behavior, like reaching a peak or a valley. For the function \(f(x)=\sqrt{4-x^{2}}\), the critical point analysis showed that there are no critical points except the endpoints, due to the nature of the derivative.
Derivative
The derivative of a function gives us the slope or rate of change at any given point. In simple terms, it tells us how the function is behaving: increasing, decreasing, or staying flat. Derivatives are indispensable in calculus, especially when it comes to finding extrema points.
  • Use derivative to check for increasing/decreasing functions
  • Find points where derivative is zero for potential extrema
For the given function, the derivative is found using the chain rule, resulting in \(f'(x) = -\frac{x}{\sqrt{4-x^{2}}}\). This derivative helps identify potential critical points, which in turn help locate where the extrema might occur within a given interval.
Intervals
In calculus, an interval refers to a specific range of x-values over which we examine a function. Calculating extrema over different intervals can yield different results, emphasizing the importance of context when analyzing functions. An interval can be:
  • Closed: Includes its endpoints, noted by brackets \([a, b]\)
  • Open: Does not include endpoints, noted by parentheses \((a, b)\)
  • Half-open: Includes only one endpoint, e.g., \([a, b)\) or \((a, b]\)
Each interval is unique in that the presence or absence of endpoints changes how we evaluate the function. For instance, an absolute maximum or minimum can occur at an endpoint, but this possibility vanishes when endpoints are excluded, such as in an open interval.

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Most popular questions from this chapter

The cross sections of an irrigation canal are isosceles trapezoids of which three sides are 8 feet long (see figure). Determine the angle of elevation \(\theta\) of the sides such that the area of the cross section is a maximum by completing the following. (a) Analytically complete six rows of a table such as the one below. (The first two rows are shown.) $$ \begin{array}{|c|c|c|c|} \hline \text { Base 1 } & \text { Base 2 } & \text { Altitude } & \text { Area } \\ \hline 8 & 8+16 \cos 10^{\circ} & 8 \sin 10^{\circ} & \approx 22.1 \\ \hline 8 & 8+16 \cos 20^{\circ} & 8 \sin 20^{\circ} & \approx 42.5 \\ \hline \end{array} $$ (b) Use a graphing utility to generate additional rows of the table and estimate the maximum cross-sectional area. (c) Write the cross-sectional area \(A\) as a function of \(\theta\). (d) Use calculus to find the critical number of the function in part (c) and find the angle that will yield the maximum cross-sectional area. (e) Use a graphing utility to graph the function in part (c) and verify the maximum cross-sectional area.

Consider \(\lim _{x \rightarrow \infty} \frac{3 x}{\sqrt{x^{2}+3}}\). Use the definition of limits at infinity to find values of \(M\) that correspond to (a) \(\varepsilon=0.5\) and (b) \(\varepsilon=0.1\).

In Exercises 87 and \(88,\) (a) use a graphing utility to graph \(f\) and \(g\) in the same viewing window, (b) verify algebraically that \(f\) and \(g\) represent the same function, and (c) zoom out sufficiently far so that the graph appears as a line. What equation does this line appear to have? (Note that the points at which the function is not continuous are not readily seen when you zoom out.) $$ \begin{array}{l} f(x)=\frac{x^{3}-3 x^{2}+2}{x(x-3)} \\ g(x)=x+\frac{2}{x(x-3)} \end{array} $$

Use the Intermediate Value Theorem and Rolle's Theorem to prove that the equation has exactly one real Solution. $$ 2 x-2-\cos x=0 $$

Numerical, Graphical, and Analytic Analysis Consider the functions \(f(x)=x\) and \(g(x)=\tan x\) on the interval \((0, \pi / 2)\) (a) Complete the table and make a conjecture about which is the greater function on the interval \((0, \pi / 2)\). $$ \begin{array}{|l|l|l|l|l|l|l|} \hline x & 0.25 & 0.5 & 0.75 & 1 & 1.25 & 1.5 \\ \hline f(x) & & & & & & \\ \hline g(x) & & & & & & \\ \hline \end{array} $$ (b) Use a graphing utility to graph the functions and use the graphs to make a conjecture about which is the greater function on the interval \((0, \pi / 2)\). (c) Prove that \(f(x)0,\) where \(h=g-f .\)
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