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Chapter 1: Problem 74

Graph the function and determine the interval(s) for which \(f(x) \geq 0\). $$ f(x)=\frac{1}{2}(2+|x|) $$

Short Answer

Expert verified
The function \(f(x)\) is greater than or equal to zero for all real numbers, in other words, the interval is \(-\infty < x < \infty\).

Step by step solution

01

Identify the base function

The function \(f(x)=\frac{1}{2}(2+|x|)\) is a transformation of the base function \(|x|\). In general, \(|x|\) returns the absolute value of \(x\), which is \(x\) if \(x \geq 0\), and \(-x\) if \(x < 0\). Therefore, our function \(f(x)\) will split into two cases based on these conditions.
02

Define two cases

Given \(f(x) = \frac{1}{2} (2 + |x|)\), if \(x \geq 0\), the function is \(f(x) = \frac{1}{2} (2 + x)\), simplifying yields \(f(x) = x + 1\). If \(x < 0\), \(f(x) = \frac{1}{2}(2 - x)\) which simplifies to \(f(x) = 1 - \frac{1}{2}x\). These two functions will help us in graphing the function and finding the intervals where \(f(x) \geq 0\).
03

Graph the function

First, graph the two equations found in the previous step. The graph of \(f(x) = x + 1\) will intersect the y-axis at \(y=1\) and have a slope of 1, shown for \(x \geq 0\). The graph of the equation \(f(x) = 1 - \frac{1}{2}x \) will intersect the y-axis at \(y=1\) but with a slope of \(-\frac{1}{2}\), shown for \(x < 0\). Both graphs will meet at the point \((0, 1)\)
04

Determine the intervals

Looking at the graph, we can see that the function \(f(x)\) is always greater than or equal to 0. This is because the function always delivers a positive value or zero. So the interval for which \(f(x) \geq 0\) is \(-\infty < x < \infty\).

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