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Chapter 1: Problem 83

\(f(x)=\max \\{2-x, 2+x, 4\\} x \in R\) is (a) \(f(x)=\mid \begin{array}{cc}2-x & x \geq 2 \\ 4 & -2

Short Answer

Expert verified
The correct piecewise definition for the given function is: \(f(x) = \left\{ \begin{array}{cc} 2+x & x \leq -2 \\ 4 & -2 < x < 2 \\ 2-x & x \geq 2 \end{array} \right. \)

Step by step solution

01

Check the endpoints of the intervals

We have the intervals: \(x \leq -2, -2 < x < 2,\) and \(x \geq 2\). Let's check the endpoints \(x = -2\) and \(x = 2\). For \(x = -2\): \( 2-x = 2-(-2) = 4 \\ 2+x = 2+(-2) = 0 \\ \) For \(x = 2\): \( 2-x = 2-2 = 0 \\ 2+x = 2+2 = 4 \\ \)
02

Determine the correct function forms for each interval

Now, let's determine the highest function form for each interval. \(x \leq -2\): Considering the function values at the endpoint \(x = -2\), we saw that \(2-x = 4 = \max \{4, 0\}\). Therefore, the function form in this interval is \(2+x\). \(-2 < x < 2\): In this interval, the function forms \(2-x\) and \(2+x\) are both increasing from their respective endpoints, so they cannot be the highest simultaneously. As \(4\) is constant, we know that the maximum value in this interval is \(f(x) = 4\). \(x \geq 2\): Considering the function values at the endpoint \(x = 2\), we saw that \(2+x = 4 = \max \{0,4\}\). Therefore, the function form in this interval is \(2-x\).
03

Find the correct match among the given options

Now that we have found the correct function forms for each interval, we can look at the given options and determine which one matches our result. Our result: \( f(x) = \left\{ \begin{array}{cc} 2+x & x \leq -2 \\ 4 & -2 < x < 2 \\ 2-x & x \geq 2 \end{array} \right. \) Comparing it with the given options, we can see that option (d) is the correct one. So, the correct piecewise definition for the function is: \(f(x) = \left\{ \begin{array}{cc} 2-x & x \leq -2 \\ 4 & -2 < x < 2 \\ 2+x & x \geq 2 \end{array} \right. \)

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