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Expert-verified Found in: Page 209 ### Essential Calculus: Early Transcendentals

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 $$f(0) = - 1$$.

See the 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 $$f(0) = - 1$$. ### Want to see more solutions like these? 