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Essential Calculus: Early Transcendentals
Found in: Page 209
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Sketch the graph of \(f\) by hand and use your sketch to find the absolute and local maximum and minimum values of \(f\). (Use the graphs and transformations of Sections 1.2.)

22. \(f(x) = \left\{ {\begin{aligned}{*{20}{l}}{4 - {x^2}}&{ if - 2 \le x < 0}\\{2x - 1}&{ if 0 \le x \le 2}\end{aligned}} \right.\)

The absolute minimum is \(f(0) = - 1\).

See the step by step solution

Step by Step Solution

Step 1: Given data

The given function is \(f(x)\).

Step 2: Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Step 3: Sketch the graph of the function

Let the given function be \(y\). Then, .

\(f(x) = \left\{ {\begin{aligned}{*{20}{l}}{4 - {x^2}}&{{\rm{ if }} - 2 \le x < 0}\\{2x - 1}&{{\rm{ if }}0 \le x \le 2}\end{aligned}} \right.\)

Use the function \(f(x) = 4 - {x^2}\) if \( - 2 \le x < 0\) and \(f(x) = 2x - 1\) if \(0 \le x \le 2\).

From Figure 1, it is observe that \(f(0) \le f(x)\) for all the values of \(x\).

Therefore, the absolute minimum is \(f(0) = - 1\). That is, the absolute minimum occurs at \(x = 0\). Note that, it has no maximum value as the function takes on values arbitrarily close to 4 but does not attain 4. Therefore, the absolute minimum is \(f(0) = - 1\).

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