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The Absolute Value Function, and its Properties

The Absolute Value Function, and its PropertiesOne of the most used functions in mathematics is the Absolute Value function. Its definition andsome of its Properties are given Value FunctionThe Absolute Value of a real numberx,|x|, is|x|={xifx 0 xifx <0 The graph of the Absolute Value function is shown belowxyExample 1|2|= 2,| 2|= ( 2) = 2 The Absolute Value function is used to measure the distance between two numbers. Thus, thedistance betweenxand 0 is|x 0|=|x|, and the distance betweenxandyis|x y|. Thus, thedistance from 2 to 4 is| 2 ( 4)|=| 2 + 4|=|2|= 2, and the distance from 2 to 5 is| 2 5|=| 7|= following Properties of the Absolute Value function need to be any two real numbersxandy, we have|xy|=|x| |y|.}

The following properties of the absolute value function need to be memorized. Lemma 1.For any two real numbers x and y, we have jxyj= jxjjyj. This equality can be veri ed by considering cases. One of the four possible cases is checked as follows: Suppose x < 0 and y 0. Then xy is 0 and we have jxyj= (xy) = ( x)y = jxjjyj.

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Transcription of The Absolute Value Function, and its Properties

1 The Absolute Value Function, and its PropertiesOne of the most used functions in mathematics is the Absolute Value function. Its definition andsome of its Properties are given Value FunctionThe Absolute Value of a real numberx,|x|, is|x|={xifx 0 xifx <0 The graph of the Absolute Value function is shown belowxyExample 1|2|= 2,| 2|= ( 2) = 2 The Absolute Value function is used to measure the distance between two numbers. Thus, thedistance betweenxand 0 is|x 0|=|x|, and the distance betweenxandyis|x y|. Thus, thedistance from 2 to 4 is| 2 ( 4)|=| 2 + 4|=|2|= 2, and the distance from 2 to 5 is| 2 5|=| 7|= following Properties of the Absolute Value function need to be any two real numbersxandy, we have|xy|=|x| |y|.}

2 This equality can be verified by considering cases. One of the four possible cases is checkedas follows: Supposex <0 andy 0. Thenxyis 0 and we have|xy|= (xy) = ( x)y=|x| |y|.The other three cases are similarly 2 For any real numberx, and any nonnegative numbera, we have|x| aif and only if a x is verified by considering the two possible casesx 0 andx <0. We consider the caseofx 0. So supposex 0 and|x| a. Then we have a 0 x=|x| is, if|x| a, then a x a. Conversely supposex 0 and a x a. Then we have|x|=x argument for the casex <0 is similar, and left to the 3 For any real numberx, and any nonnegative numbera, we have|x| aif and only ifx aorx are again two cases to consider:xpositive orxnegative.

3 So supposex 0. If|x| a, then we havex=|x| aOn the other hand ifx a, then we have|x|=x a. Conversely supposex <0 and|x| a,then we have x=|x| ax the case wherex <0, if we havex a, then|x|= x ( a) =aNote whenx <0 the conditionx acannot be true, since we are assumingx 4 For any two real numbersxandywe have|x+y| |x|+|y|.The proof of this is again handled by considering the four possible cases determined by thesigns ofxandy. The simplest case being when bothxandyare nonnegative; in this case we have|x+y|=x+y=|x|+|y|The other three cases are similarly dealt : in any of the above inequalities the less than or equal relation can be replaced with strictinequality.

4 For example,|x|< aif and only if a < x < 2 Find the interval of real numbers which containsx, ifxsatisfies the condition|2x 5|<3.|2x 5|<3 3<2x 5<32<2x <81< x <9 .Example 3 How close must the numberxbe to 4 if|3x 12|<5.|3x 12|<5.|3 (x 4)|<53|x 4|<5|x 4|<5/3 Thus,xmust be within 5/3 of 4 if 3xis to be within 5 units of 12.


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