. Absolute Value Functions 7. Absolute Value Functions There are a few was to describe what is meant b the absolute value of a real number. You ma have been taught that is the distance from the real number to 0 on the number line. So, for eample, 5 = 5 and 5 = 5, since each is 5 units from 0 on the number line. distance is 5 units distance is 5 units 5 0 5 Another wa to define absolute value is b the equation =. Using this definition, we have 5 = (5) = 5 = 5 and 5 = ( 5) = 5 = 5. The long and short of both of these procedures is that takes negative real numbers and assigns them to their positive counterparts while it leaves positive numbers alone. This last description is the one we shall adopt, and is summarized in the following definition. Definition.. The absolute value of a real number, denoted, is given b, if < 0 =, if 0 In Definition., we define using a piecewise-defined function. (See page 6 in Section..) To check that this definition agrees with what we previousl understood as absolute value, note that since 5 0, to find 5 we use the rule =, so 5 = 5. Similarl, since 5 < 0, we use the rule =, so that 5 = ( 5) = 5. This is one of the times when it s best to interpret the epression as the opposite of as opposed to negative. Before we begin studing absolute value functions, we remind ourselves of the properties of absolute value. Theorem.. Properties of Absolute Value: Let a, b and be real numbers and let n be an integer. a Then Product Rule: ab = a b Power Rule: a n = a n whenever a n is defined Quotient Rule: a = a b b, provided b 0 Equalit Properties: = 0 if and onl if = 0. For c > 0, = c if and onl if = c or = c. For c < 0, = c has no solution. a See page if ou don t remember what an integer is.
7 Linear and Quadratic Functions The proofs of the Product and Quotient Rules in Theorem. boil down to checking four cases: when both a and b are positive; when the are both negative; when one is positive and the other is negative; and when one or both are zero. For eample, suppose we wish to show that ab = a b. We need to show that this equation is true for all real numbers a and b. If a and b are both positive, then so is ab. Hence, a = a, b = b and ab = ab. Hence, the equation ab = a b is the same as ab = ab which is true. If both a and b are negative, then ab is positive. Hence, a = a, b = b and ab = ab. The equation ab = a b becomes ab = ( a)( b), which is true. Suppose a is positive and b is negative. Then ab is negative, and we have ab = ab, a = a and b = b. The equation ab = a b reduces to ab = a( b) which is true. A smmetric argument shows the equation ab = a b holds when a is negative and b is positive. Finall, if either a or b (or both) are zero, then both sides of ab = a b are zero, so the equation holds in this case, too. All of this rhetoric has shown that the equation ab = a b holds true in all cases. The proof of the Quotient Rule is ver similar, with the eception that b 0. The Power Rule can be shown b repeated application of the Product Rule. The Equalit Properties can be proved using Definition. and b looking at the cases when 0, in which case =, or when < 0, in which case =. For eample, if c > 0, and = c, then if 0, we have = = c. If, on the other hand, < 0, then = = c, so = c. The remaining properties are proved similarl and are left for the Eercises. Our first eample reviews how to solve basic equations involving absolute value using the properties listed in Theorem.. Eample... Solve each of the following equations.. = 6. + 5 =. + 5 = 0. 5 + = 5 5. = 6 6. + = Solution.. The equation = 6 is of the form = c for c > 0, so b the Equalit Properties, = 6 is equivalent to = 6 or = 6. Solving the former, we arrive at = 7, and solving the latter, we get = 5. We ma check both of these solutions b substituting them into the original equation and showing that the arithmetic works out.. To use the Equalit Properties to solve + 5 =, we first isolate the absolute value. + 5 = + 5 = subtract + 5 = divide b From the Equalit Properties, we have + 5 = or + 5 =, and get our solutions to be = or = 7. We leave it to the reader to check both answers in the original equation.
. Absolute Value Functions 75. As in the previous eample, we first isolate the absolute value in the equation + 5 = 0 and get + = 5. Using the Equalit Properties, we have + = 5 or + = 5. Solving the former gives = and solving the latter gives =. As usual, we ma substitute both answers in the original equation to check.. Upon isolating the absolute value in the equation 5 + = 5, we get 5 + =. At this point, we know there cannot be an real solution, since, b definition, the absolute value of anthing is never negative. We are done. 5. The equation = 6 presents us with some difficult, since appears both inside and outside of the absolute value. Moreover, there are values of for which 6 is positive, negative and zero, so we cannot use the Equalit Properties without the risk of introducing etraneous solutions, or worse, losing solutions. For this reason, we break equations like this into cases b rewriting the term in absolute values,, using Definition.. For < 0, =, so for < 0, the equation = 6 is equivalent to = 6. Rearranging this gives us + 6 = 0, or (+)( ) = 0. We get = or =. Since onl = satisfies < 0, this is the answer we keep. For 0, =, so the equation = 6 becomes = 6. From this, we get 6 = 0 or ( )( + ) = 0. Our solutions are = or =, and since onl = satisfies 0, this is the one we keep. Hence, our two solutions to = 6 are = and =. 6. To solve + =, we first isolate the absolute value and get =. Since we see both inside and outside of the absolute value, we break the equation into cases. The term with absolute values here is, so we replace with the quantit ( ) in Definition. to get ( ), if ( ) < 0 = ( ), if ( ) 0 Simplifing ields = +, if <, if So, for <, = + and our equation = becomes + =, which gives =. Since this solution satisfies <, we keep it. Net, for, =, so the equation = becomes =. Here, the equation reduces to =, which signifies we have no solutions here. Hence, our onl solution is =. Net, we turn our attention to graphing absolute value functions. Our strateg in the net eample is to make liberal use of Definition. along with what we know about graphing linear functions (from Section.) and piecewise-defined functions (from Section.). Eample... Graph each of the following functions.. f() =. g() =. h() =. i() = +
76 Linear and Quadratic Functions Find the zeros of each function and the - and -intercepts of each graph, if an eist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute etrema, if the eist. Solution.. To find the zeros of f, we set f() = 0. We get = 0, which, b Theorem. gives us = 0. Since the zeros of f are the -coordinates of the -intercepts of the graph of = f(), we get (0, 0) as our onl -intercept. To find the -intercept, we set = 0, and find = f(0) = 0, so that (0, 0) is our -intercept as well. Using Definition., we get f() = =, if < 0, if 0 Hence, for < 0, we are graphing the line = ; for 0, we have the line =. Proceeding as we did in Section.6, we get f() =, < 0 f() =, 0 Notice that we have an open circle at (0, 0) in the graph when < 0. As we have seen before, this is due to the fact that the points on = approach (0, 0) as the -values approach 0. Since is required to be strictl less than zero on this stretch, the open circle is drawn at the origin. However, notice that when 0, we get to fill in the point at (0, 0), which effectivel plugs the hole indicated b the open circle. Thus we get, f() = Actuall, since functions can have at most one -intercept (Do ou know wh?), as soon as we found (0, 0) as the -intercept, we knew this was also the -intercept.
. Absolute Value Functions 77 B projecting the graph to the -ais, we see that the domain is (, ). Projecting to the -ais gives us the range [0, ). The function is increasing on [0, ) and decreasing on (, 0]. The relative minimum value of f is the same as the absolute minimum, namel 0 which occurs at (0, 0). There is no relative maimum value of f. There is also no absolute maimum value of f, since the values on the graph etend infinitel upwards.. To find the zeros of g, we set g() = = 0. B Theorem., we get = 0 so that =. Hence, the -intercept is (, 0). To find our -intercept, we set = 0 so that = g(0) = 0 =, which ields (0, ) as our -intercept. To graph g() =, we use Definition. to rewrite g as g() = = ( ), if ( ) < 0 ( ), if ( ) 0 Simplifing, we get g() = +, if <, if As before, the open circle we introduce at (, 0) from the graph of = + is filled b the point (, 0) from the line =. We determine the domain as (, ) and the range as [0, ). The function g is increasing on [, ) and decreasing on (, ]. The relative and absolute minimum value of g is 0 which occurs at (, 0). As before, there is no relative or absolute maimum value of g.. Setting h() = 0 to look for zeros gives = 0. As in Eample.., we isolate the absolutre value to get = so that = or =. As a result, we have a pair of - intercepts: (, 0) and (, 0). Setting = 0 gives = h(0) = 0 =, so our -intercept is (0, ). As before, we rewrite the absolute value in h to get h() =, if < 0, if 0 Once again, the open circle at (0, ) from one piece of the graph of h is filled b the point (0, ) from the other piece of h. From the graph, we determine the domain of h is (, ) and the range is [, ). On [0, ), h is increasing; on (, 0] it is decreasing. The relative minimum occurs at the point (0, ) on the graph, and we see is both the relative and absolute minimum value of h. Also, h has no relative or absolute maimum value.
78 Linear and Quadratic Functions 5 g() = h() =. As before, we set i() = 0 to find the zeros of i and get + = 0. Isolating the absolute value term gives + =, so either + = or + =. We get = or =, so our -intercepts are (, 0) and (, 0). Substituting = 0 gives = i(0) = (0)+ =, for a -intercept of (0, ). Rewriting the formula for i() without absolute values gives i() = ( ( + )), if ( + ) < 0 6 + 6, if < ( + ), if ( + ) 0 = 6 +, if The usual analsis near the trouble spot = gives the corner of this graph is (, ), and we get the distinctive shape: 5 i() = + The ( domain ] of i is (, ) [ while the range is (, ]. The function i is increasing on, and decreasing on, ). The relative maimum occurs at the point (, ) and the relative and absolute maimum value of i is. Since the graph of i etends downwards forever more, there is no absolute minimum value. As we can see from the graph, there is no relative minimum, either. Note that all of the functions in the previous eample bear the characteristic shape of the graph of =. We could have graphed the functions g, h and i in Eample.. starting with the graph of f() = and appling transformations as in Section.7 as our net eample illustrates.
. Absolute Value Functions 79 Eample... Graph the following functions starting with the graph of f() = and using transformations.. g() =. h() =. i() = + Solution. We begin b graphing f() = and labeling three points, (, ), (0, 0) and (, ). (, ) (, ) (0, 0) f() =. Since g() = = f( ), Theorem.7 tells us to add to each of the -values of the points on the graph of = f() to obtain the graph of = g(). This shifts the graph of = f() to the right units and moves the point (, ) to (, ), (0, 0) to (, 0) and (, ) to (, ). Connecting these points in the classic fashion produces the graph of = g(). (, ) (, ) (, ) (, ) (0, 0) f() = shift right units add to each -coordinate (, 0) 5 6 g() = f( ) =. For h() = = f(), Theorem.7 tells us to subtract from each of the -values of the points on the graph of = f() to obtain the graph of = h(). This shifts the graph of = f() down units and moves (, ) to (, ), (0, 0) to (0, ) and (, ) to (, ). Connecting these points with the shape produces our graph of = h(). (, ) (, ) (, ) (, ) (0, ) (0, 0) f() = shift down units subtract from each -coordinate h() = f() =
80 Linear and Quadratic Functions. We re-write i() = + = f( + ) = f( + ) + and appl Theorem.7. First, we take care of the changes on the inside of the absolute value. Instead of, we have +, so, in accordance with Theorem.7, we first subtract from each of the -values of points on the graph of = f(), then divide each of those new values b. This effects a horizontal shift left unit followed b a horizontal shrink b a factor of. These transformations move (, ) to (, ), (0, 0) to (, 0) and (, ) to (0, ). Net, we take care of what s happening outside of the absolute value. Theorem.7 instructs us to first multipl each -value of these new points b then add. Geometricall, this corresponds to a vertical stretch b a factor of, a reflection across the -ais and finall, a vertical shift up units. These transformations move (, ) to (, ), (, 0) to (, ), and (0, ) to (0, ). Connecting these points with the usual shape produces our graph of = i(). (, ) (0, 0) (, ) `, `, (0, ) f() = i() = f( + ) + = + + While the methods in Section.7 can be used to graph an entire famil of absolute value functions, not all functions involving absolute values posses the characteristic shape. As the net eample illustrates, often there is no substitute for appealing directl to the definition. Eample... Graph each of the following functions. Find the zeros of each function and the - and -intercepts of each graph, if an eist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute etrema, if the eist.. f() =. g() = + + Solution.. We first note that, due to the fraction in the formula of f(), 0. Thus the domain is (, 0) (0, ). To find the zeros of f, we set f() = = 0. This last equation implies = 0, which, from Theorem., implies = 0. However, = 0 is not in the domain of f,
. Absolute Value Functions 8 which means we have, in fact, no -intercepts. We have no -intercepts either, since f(0) is undefined. Re-writing the absolute value in the function gives f() =, if < 0 =, if > 0, if < 0, if > 0 To graph this function, we graph two horizontal lines: = for < 0 and = for > 0. We have open circles at (0, ) and (0, ) (Can ou eplain wh?) so we get f() = As we found earlier, the domain is (, 0) (0, ). The range consists of just two -values:, }. The function f is constant on (, 0) and (0, ). The local minimum value of f is the absolute minimum value of f, namel ; the local maimum and absolute maimum values for f also coincide the both are. Ever point on the graph of f is simultaneousl a relative maimum and a relative minimum. (Can ou remember wh in light of Definition.? This was eplored in the Eercises in Section.6..). To find the zeros of g, we set g() = 0. The result is + + = 0. Attempting to isolate the absolute value term is complicated b the fact that there are two terms with absolute values. In this case, it easier to proceed using cases b re-writing the function g with two separate applications of Definition. to remove each instance of the absolute values, one at a time. In the first round we get g() = ( + ) +, if ( + ) < 0 ( + ) +, if ( + ) 0 =, if < +, if Given that = ( ), if ( ) < 0, if ( ) 0 = +, if <, if, we need to break up the domain again at =. Note that if <, then <, so we replace with + for that part of the domain, too. Our completed revision of the form of g ields
8 Linear and Quadratic Functions g() = ( + ), if < + ( + ), if and < + ( ), if =, if <, if < 6, if To solve g() = 0, we see that the onl piece which contains a variable is g() = for <. Solving = 0 gives = 0. Since = 0 is in the interval [, ), we keep this solution and have (0, 0) as our onl -intercept. Accordingl, the -intercept is also (0, 0). To graph g, we start with < and graph the horizontal line = with an open circle at (, ). For <, we graph the line = and the point (, ) patches the hole left b the previous piece. An open circle at (, 6) completes the graph of this part. Finall, we graph the horizontal line = 6 for, and the point (, 6) fills in the open circle left b the previous part of the graph. The finished graph is 6 5 g() = + + The domain of g is all real numbers, (, ), and the range of g is all real numbers between and 6 inclusive, [, 6]. The function is increasing on [, ] and constant on (, ] and [, ). The relative minimum value of f is which matches the absolute minimum. The relative and absolute maimum values also coincide at 6. Ever point on the graph of = g() for < and > ields both a relative minimum and relative maimum. The point (, ), however, gives onl a relative minimum and the point (, 6) ields onl a relative maimum. (Recall the Eercises in Section.6. which dealt with constant functions.) Man of the applications that the authors are aware of involving absolute values also involve absolute value inequalities. For that reason, we save our discussion of applications for Section..
. Absolute Value Functions 8.. Eercises In Eercises - 5, solve the equation.. = 6. = 0. = 7. = 5. 5 + = 0 6. 7 + = 0 7. 5 = 8. 5 = 5 9. = + 0. = +. = +. = 5. =. = 5. = Prove that if f() = g() then either f() = g() or f() = g(). Use that result to solve the equations in Eercises 6 -. 6. = + 7 7. + = 8. = + 9. + = 0 0. 5 = 5 +. = + In Eercises -, graph the function. Find the zeros of each function and the - and -intercepts of each graph, if an eist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute etrema, if the eist.. f() = +. f() = +. f() = 5. f() = 6. f() = + 7. f() = 8. f() = + + 9. f() = 0. f() = +. f() = +. f() = +. f() = + +. With the help of our classmates, find an absolute value function whose graph is given below. 8 7 6 5 5 6 7 8 5. With help from our classmates, prove the second, third and fifth parts of Theorem.. 6. Prove The Triangle Inequalit: For all real numbers a and b, a + b a + b.
8 Linear and Quadratic Functions.. Answers. = 6 or = 6. = or =. = or =. = or = 5. = or = 0 6. no solution 7. = or = 8. = 8 or = 5 8 9. = 0. = 0 or =. =. no solution. =, = 0 or =. = or = 5. = or = 6. = or = 9 7. = 7 or = 8. = 0 or = 9. = 0. = 0. = 5 or = 5. f() = + f( ) = 0 -intercept (, 0) -intercept (0, ) Range [0, ) Decreasing on (, ] Increasing on [, ) Relative and absolute min. at (, 0) No relative or absolute maimum 8 7 6 5. f() = + No zeros No -intercepts -intercept (0, ) Range [, ) Decreasing on (, 0] Increasing on [0, ) Relative and absolute minimum at (0, ) No relative or absolute maimum 8 7 6 5
. Absolute Value Functions 85. f() = f(0) = 0 -intercept (0, 0) -intercept (0, 0) Range [0, ) Decreasing on (, 0] Increasing on [0, ) Relative and absolute minimum at (0, 0) No relative or absolute maimum 8 7 6 5 5. f() = f(0) = 0 -intercept (0, 0) -intercept (0, 0) Range (, 0] Increasing on (, 0] Decreasing on [0, ) Relative and absolute maimum at (0, 0) No relative or absolute minimum 5 6 6. f() = + f ( 6 ) ( = 0, f 8 ) = 0 -intercepts ( 6, 0), ( 8, 0) -intercept (0, 8) Range [, ) Decreasing on (, ] Increasing on [, ) Relative and absolute min. at (, ) No relative or absolute maimum 8 7 6 5 8 7 6 5 7. f() = f ( ) = 0 -intercepts (, 0) -intercept ( 0, ) Range [0, ) Decreasing on (, ] Increasing on [, ) Relative and absolute min. at (, 0) No relative or absolute maimum
86 Linear and Quadratic Functions + 8. f() = + No zeros No -intercept -intercept (0, ) Domain (, ) (, ) Range, } Constant on (, ) Constant on (, ) Absolute minimum at ever point (, ) 9. f() = No zeros No -intercept -intercept (0, ) Domain (, ) (, ) Range, } Constant on (, ) Constant on (, ) Absolute minimum at ever point (, ) 0. Re-write f() = + as if < 0 f() = if 0 f ( ) = 0 -intercept (, 0) -intercept (0, ) Range [, ) Increasing on [0, ) Constant on (, 0] Absolute minimum at ever point (, ) where 0 No absolute maimum where < Absolute maimum at ever point (, ) where > Relative maimum AND minimum at ever point on the graph 8 7 6 5 where > Absolute maimum at ever point (, ) where < Relative maimum AND minimum at ever point on the graph 5 Relative minimum at ever point (, ) where 0 Relative maimum at ever point (, ) where < 0
. Absolute Value Functions 87. Re-write f() = + as if < f() = if No zeros No -intercepts -intercept (0, ) Range [, ) Decreasing on (, ] Constant on [, ) Absolute minimum at ever point (, ) where. Re-write f() = + as if < f() = + if < 0 if 0 f ( ) = 0 -intercept (, 0) -intercept (0, ) Range [, ] Increasing on [, 0] Constant on (, ] Constant on [0, ) Absolute minimum at ever point (, ) where. Re-write f() = + + as if < f() = 6 if < + if No zeros No -intercept -intercept (0, 6) Range [6, ) Decreasing on (, ] Constant on [, ] Increasing on [, ) Absolute minimum at ever point (, 6) where No absolute maimum Relative minimum at ever point (, 6) where No absolute maimum Relative minimum at ever point (, ) where Relative maimum at ever point (, ) where > Absolute maimum at ever point (, ) where 0 Relative minimum at ever point (, ) where and at ever point (, ) where > 0 Relative maimum at ever point (, ) where < and at ever point (, ) where 0 Relative maimum at ever point (, 6) where < < 5 8 7 6 5 5. f() =