Absolute Value functions => Piecewise Function

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To express the function f(x) = |x + 1| + |x + 2| as a piecewise function, consider the critical points where each absolute value expression changes sign, specifically at x = -1 and x = -2. The function can be divided into intervals based on these points: when both expressions are positive (x ≥ -1), the function simplifies to 2x + 3. For the intervals where one or both expressions are negative, the corresponding piecewise expressions can be derived by analyzing the signs of (x + 1) and (x + 2). It's important to follow the convention that |x| = x for x ≥ 0, though both forms yield the same function. Using a table of values can also aid in determining the correct piecewise segments.
danago
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Given the function:

<br /> f(x) = \left| {x + 1} \right| + \left| {x + 2} \right|<br />

How can i write that as a piecewise function? If i was given something in the form of f(x) = \left| {g(x)} \right|, i know to write it as:

<br /> f(x) = \left\{ {\begin{array}{*{20}c}<br /> {g(x),} &amp; {g(x) \ge 0} \\<br /> { - g(x),} &amp; {g(x) &lt; 0} \\<br /> \end{array}} \right.<br /> <br />

but I am a little stuck with the other one.

Thanks in advance,
Dan.
 
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consider the cases, x+1<0, x+2<0 and other combinations separately:

x+1<0, x+2<0
x+1<0, x+2>0
x+1>0, x+2<0
x+1>0, x+2>0

notice that some of these conditions are impossible.
 
Hmm ok. So when both are positive x + 1,x + 2 \ge 0, the piece is (x + 1) + (x + 2) = 2x + 3. They are both positive only when x \ge - 1, so my piece (2x + 3) would only be existent over the domain x \ge - 1? Is that the way i should go about it?
 
yes, that should be correct.
 
Ok thanks for that :)
 
Just one more thing. Using <br /> y = \left| x \right| for example. At x=0, it could go either way. y could equal x or -x, so either of the following could be true:

<br /> \begin{array}{l}<br /> f(x) = \left\{ {\begin{array}{*{20}c}<br /> {x,} &amp; {x \ge 0} \\<br /> { - x,} &amp; {x &lt; 0} \\<br /> \end{array}} \right. \\ <br /> f(x) = \left\{ {\begin{array}{*{20}c}<br /> {x,} &amp; {x &gt; 0} \\<br /> { - x,} &amp; {x \le 0} \\<br /> \end{array}} \right. \\ <br /> \end{array}<br />

Does it matter which way i do it? Or is there some convention i should follow?
 
The convention is that |x|= x for x\ge 0 but, in fact, either that or |x|= -x for x\le 0 give the same function.
 
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Ok thanks.
 
Another way to do it is to find your critical points, (x=-1 and x=-2 in this case) and create an x, f(x) table to test points outside of and between those points within the function. Then you can take the points in the three regions and find the equation of the line between them which is the equation that satisfies the x value of the region it was determined from. A calculator that can come up with a table of values for you really helps with this.
 

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