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Q22E

Expert-verifiedFound in: Page 209

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**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 \(f(0) = - 1\).

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

**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.**

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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