2.4 Inequalities with Absolute Value and Quadratic Functions

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1 08 Linear and Quadratic Functions. Inequalities with Absolute Value and Quadratic Functions In this section, not onl do we develop techniques for solving various classes of inequalities analticall, we also look at them graphicall. The first eample motivates the core ideas. Eample... Let f() = and g() =.. Solve f() = g().. Solve f() < g().. Solve f() > g().. Graph = f() and = g() on the same set of aes and interpret our solutions to parts through above. Solution.. To solve f() = g(), we replace f() with and g() with to get =. Solving for, we get =.. The inequalit f() < g() is equivalent to <. Solving gives < or (, ).. To find where f() > g(), we solve >. We get >, or (, ).. To graph = f(), we graph =, which is a line with a -intercept of (0, ) and a slope of. The graph of = g() is = which is a horizontal line through (0, ) = g() = f() To see the connection between the graph and the Algebra, we recall the Fundamental Graphing Principle for Functions in Section.6: the point (a, b) is on the graph of f if and onl if f(a) = b. In other words, a generic point on the graph of = f() is (, f()), and a generic

2 . Inequalities with Absolute Value and Quadratic Functions 09 point on the graph of = g() is (, g()). When we seek solutions to f() = g(), we are looking for values whose values on the graphs of f and g are the same. In part, we found = is the solution to f() = g(). Sure enough, f() = and g() = so that the point (, ) is on both graphs. In other words, the graphs of f and g intersect at (, ). In part, we set f() < g() and solved to find <. For <, the point (, f()) is below (, g()) since the values on the graph of f are less than the values on the graph of g there. Analogousl, in part, we solved f() > g() and found >. For >, note that the graph of f is above the graph of g, since the values on the graph of f are greater than the values on the graph of g for those values of = f() 6 = g() 6 = g() = f() f() < g() on (, ) f() > g() on (, ) The preceding eample demonstrates the following, which is a consequence of the Fundamental Graphing Principle for Functions. Graphical Interpretation of Equations and Inequalities Suppose f and g are functions. The solutions to f() = g() are the values where the graphs of = f() and = g() intersect. The solution to f() < g() is the set of values where the graph of = f() is below the graph of = g(). The solution to f() > g() is the set of values where the graph of = f() above the graph of = g(). The net eample turns the tables and furnishes the graphs of two functions and asks for solutions to equations and inequalities.

3 0 Linear and Quadratic Functions Eample... The graphs of f and g are below. (The graph of = g() is bolded.) Use these graphs to answer the following questions. = g() (, ) (, ) = f(). Solve f() = g().. Solve f() < g().. Solve f() g(). Solution.. To solve f() = g(), we look for where the graphs of f and g intersect. These appear to be at the points (, ) and (, ), so our solutions to f() = g() are = and =.. To solve f() < g(), we look for where the graph of f is below the graph of g. This appears to happen for the values less than and greater than. Our solution is (, ) (, ).. To solve f() g(), we look for solutions to f() = g() as well as f() > g(). We solved the former equation and found = ±. To solve f() > g(), we look for where the graph of f is above the graph of g. This appears to happen between = and =, on the interval (, ). Hence, our solution to f() g() is [, ]. = g() = g() (, ) (, ) (, ) (, ) f() < g() = f() f() g() = f()

4 . Inequalities with Absolute Value and Quadratic Functions We now turn our attention to solving inequalities involving the absolute value. following theorem from Intermediate Algebra to help us. Theorem.. Inequalities Involving the Absolute Value: Let c be a real number. For c > 0, < c is equivalent to c < < c. For c > 0, c is equivalent to c c. For c 0, < c has no solution, and for c < 0, c has no solution. For c 0, > c is equivalent to < c or > c. For c 0, c is equivalent to c or c. For c < 0, > c and c are true for all real numbers. We have the As with Theorem. in Section., we could argue Theorem. using cases. However, in light of what we have developed in this section, we can understand these statements graphicall. For instance, if c > 0, the graph of = c is a horizontal line which lies above the -ais through (0, c). To solve < c, we are looking for the values where the graph of = is below the graph of = c. We know that the graphs intersect when = c, which, from Section., we know happens when = c or = c. Graphing, we get ( c, c) (c, c) c c We see that the graph of = is below = c for between c and c, and hence we get < c is equivalent to c < < c. The other properties in Theorem. can be shown similarl. Eample... Solve the following inequalities analticall; check our answers graphicall... + >. <. + + Solution.. From Theorem., is equivalent to or. Solving, we get or, which, in interval notation is (, ] [, ). Graphicall, we have

5 Linear and Quadratic Functions = = We see that the graph of = is above the horizontal line = for < and > hence this is where >. The two graphs intersect when = and =, so we have graphical confirmation of our analtic solution.. To solve + > analticall, we first isolate the absolute value before appling Theorem.. To that end, we get + > 6 or + <. Rewriting, we now have < + < so that < <. In interval notation, we write (, ). Graphicall we see that the graph of = + is above = for values between and. = + =. Rewriting the compound inequalit < as < and allows us to solve each piece using Theorem.. The first inequalit, < can be re-written as > so < or >. We get < or >. Our solution to the first inequalit is then (, ) (, ). For, we combine results in Theorems. and. to get so that 6, or [, 6]. Our solution to < is comprised of values of which satisf both parts of the inequalit, so we take intersection of (, ) (, ) and [, 6] to get [, ) (, 6]. Graphicall, we see that the graph of = is between the horizontal lines = and = for values between and as well as those between and 6. Including the values where = and = intersect, we get See Definition. in Section...

6 . Inequalities with Absolute Value and Quadratic Functions 8 = 7 6 = = We need to eercise some special caution when solving + +. As we saw in Eample.. in Section., when variables are both inside and outside of the absolute value, it s usuall best to refer to the definition of absolute value, Definition., to remove the absolute values and proceed from there. To that end, we have + = ( + ) if < and + = + if. We break the inequalit into cases, the first case being when <. For these values of, our inequalit becomes ( + ) +. Solving, we get +, so that 6, which means. Since all of these solutions fall into the categor <, we keep them all. For the second case, we assume. Our inequalit becomes + +, which gives + + or. Since all of these values of are greater than or equal to, we accept all of these solutions as well. Our final answer is (, ] [, ). = + = + We now turn our attention to quadratic inequalities. In the last eample of Section., we needed to determine the solution to 6 < 0. We will now re-visit this problem using some of the techniques developed in this section not onl to reinforce our solution in Section., but to also help formulate a general analtic procedure for solving all quadratic inequalities. If we consider f() = 6 and g() = 0, then solving 6 < 0 corresponds graphicall to finding

7 Linear and Quadratic Functions the values of for which the graph of = f() = 6 (the parabola) is below the graph of = g() = 0 (the -ais). We ve provided the graph again for reference. 6 6 = 6 We can see that the graph of f does dip below the -ais between its two -intercepts. The zeros of f are = and = in this case and the divide the domain (the -ais) into three intervals: (, ), (, ) and (, ). For ever number in (, ), the graph of f is above the -ais; in other words, f() > 0 for all in (, ). Similarl, f() < 0 for all in (, ), and f() > 0 for all in (, ). We can schematicall represent this with the sign diagram below. (+) 0 ( ) 0 (+) Here, the (+) above a portion of the number line indicates f() > 0 for those values of ; the ( ) indicates f() < 0 there. The numbers labeled on the number line are the zeros of f, so we place 0 above them. We see at once that the solution to f() < 0 is (, ). Our net goal is to establish a procedure b which we can generate the sign diagram without graphing the function. An important propert of quadratic functions is that if the function is positive at one point and negative at another, the function must have at least one zero in between. Graphicall, this means that a parabola can t be above the -ais at one point and below the -ais at another point without crossing the -ais. This allows us to determine the sign of all of the function values on a given interval b testing the function at just one value in the interval. This gives us the following. We will give this propert a name in Chapter and revisit this concept then.

8 . Inequalities with Absolute Value and Quadratic Functions Steps for Solving a Quadratic Inequalit. Rewrite the inequalit, if necessar, as a quadratic function f() on one side of the inequalit and 0 on the other.. Find the zeros of f and place them on the number line with the number 0 above them.. Choose a real number, called a test value, in each of the intervals determined in step.. Determine the sign of f() for each test value in step, and write that sign above the corresponding interval.. Choose the intervals which correspond to the correct sign to solve the inequalit. Eample... Solve the following inequalities analticall using sign diagrams. answer graphicall. Verif our.. >. +. Solution.. To solve, we first get 0 on one side of the inequalit which ields + 0. We find the zeros of f() = + b solving + = 0 for. Factoring gives ( + )( ) = 0, so = or =. We place these values on the number line with 0 above them and choose test values in the intervals (, ) (,, ) and (, ). For the interval (, ), we choose = ; for (, ), we pick = 0; and for (, ), =. Evaluating the function at the three test values gives us f( ) = > 0, so we place (+) above (, ( ) ; f(0) = < 0, so ( ) goes above the interval, ) ; and, f() = 7, which means (+) is placed above (, ). Since we are solving + 0, we look for solutions to + < 0 as well as solutions for + = 0. For + < 0, we need the intervals which we have a ( ). Checking the sign diagram, we see this is (, ). We know + = 0 when = and =, so our final answer is [, ]. To verif our solution graphicall, we refer to the original inequalit,. We let g() = and h() =. We are looking for the values where the graph of g is below that of h (the solution to g() < h()) as well as the points of intersection (the solutions to g() = h()). The graphs of g and h are given on the right with the sign chart on the left. We have to choose something in each interval. If ou don t like our choices, please feel free to choose different numbers. You ll get the same sign chart.

9 6 Linear and Quadratic Functions (+) 0 ( ) 0 (+) = =. Once again, we re-write > as > 0 and we identif f() =. When we go to find the zeros of f, we find, to our chagrin, that the quadratic doesn t factor nicel. Hence, we resort to the quadratic formula to solve = 0, and arrive at = ±. As before, these zeros divide the number line into three pieces. To help us decide on test values, we approimate 0. and +.. We choose =, = 0 and = as our test values and find f( ) =, which is (+); f(0) = which is ( ); and f() = which is (+) again. Our solution to > 0 is where we have (+), so, in interval notation (, ) ( +, ). To check the inequalit > graphicall, we set g() = and h() =. We are looking for the values where the graph of g is above the graph of h. As before we present the graphs on the right and the sign chart on the left. (+) 0 ( ) 0 (+) + 0 = =. To solve +, as before, we solve + 0. Setting f() = + = 0, we find the onl one zero of f, =. This one value divides the number line into two intervals, from which we choose = 0 and = as test values. We find f(0) = > 0 and f() = > 0. Since we are looking for solutions to + 0, we are looking for values where + < 0 as well as where + = 0. Looking at our sign diagram, there are no places where + < 0 (there are no ( )), so our solution is onl = (where + = 0). We write this as {}. Graphicall, we solve + b graphing g() = + and h() =. We are looking for the values where the graph of g is below the graph of h (for + < ) and where the two graphs intersect ( + = ). Notice that the line and the parabola touch at (, ), but the parabola is alwas above the line otherwise. In this case, we sa the line = is tangent to = + at (, ). Finding tangent lines to arbitrar functions is a fundamental problem solved, in general, with Calculus.

10 . Inequalities with Absolute Value and Quadratic Functions 7 (+) 0 (+) 0 = + =. To solve our last inequalit,, we re-write the absolute value using cases. For <, = ( ) =, so we get, or 0. Finding the zeros of f() =, we get = 0 and =. However, we are onl concerned with the portion of the number line where <, so the onl zero that we concern ourselves with is = 0. This divides the interval < into two intervals: (, 0) and (0, ). We choose = and = as our test values. We find f( ) = and f ( ) =. Hence, our solution to 0 for < is [0, ). Net, we turn our attention to the case. Here, =, so our original inequalit becomes, or 0. Setting g() =, we find the zeros of g to be = and =. Of these, onl = lies in the region, so we ignore =. Our test intervals are now [, ) and (, ). We choose = and = as our test values and find g() = and g() =. Hence, our solution to g() = 0, in this region is [, ). (+) 0 ( ) 0 ( ) 0 (+) Combining these into one sign diagram, we have that our solution is [0, ]. Graphicall, to check, we set h() = and i() = and look for the values where the graph of h is above the the graph of i (the solution of h() > i()) as well as the -coordinates of the intersection points of both graphs (where h() = i()). The combined sign chart is given on the left and the graphs are on the right. = (+) 0 ( ) 0 (+) 0 0 =

11 8 Linear and Quadratic Functions One of the classic applications of inequalities is the notion of tolerances. Recall that for real numbers and c, the quantit c ma be interpreted as the distance from to c. Solving inequalities of the form c d for d 0 can then be interpreted as finding all numbers which lie within d units of c. We can think of the number d as a tolerance and our solutions as being within an accepted tolerance of c. We use this principle in the net eample. Eample... The area A (in square inches) of a square piece of particle board which measures inches on each side is A() =. Suppose a manufacturer needs to produce a inch b inch square piece of particle board as part of a home office desk kit. How close does the side of the piece of particle board need to be cut to inches to guarantee that the area of the piece is within a tolerance of 0. square inches of the target area of 76 square inches? Solution. Mathematicall, we epress the desire for the area A() to be within 0. square inches of 76 as A Since A() =, we get 76 0., which is equivalent to One wa to proceed at this point is to solve the two inequalities and individuall using sign diagrams and then taking the intersection of the solution sets. While this wa will (eventuall) lead to the correct answer, we take this opportunit to showcase the increasing propert of the square root: if 0 a b, then a b. To use this propert, we proceed as follows (add 76 across the inequalities.) (take square roots.) ( = ) B Theorem., we find the solution to 7.7 to be (, 7.7 ] [ 7.7, ) and the solution to 76. to be [ 76., 76. ]. To solve , we intersect these two sets to get [ 76., 7.7] [ 7.7, 76.]. Since represents a length, we discard the negative answers and get [ 7.7, 76.]. This means that the side of the piece of particle board must be cut between and inches, a tolerance of (approimatel) 0.00 inches of the target length of inches. Our last eample in the section demonstrates how inequalities can be used to describe regions in the plane, as we saw earlier in Section.. Eample..6. Sketch the following relations.. R = {(, ) : > }. S = {(, ) : }. T = {(, ) : < } The underling concept of Calculus can be phrased in terms of tolerances, so this is well worth our attention.

12 . Inequalities with Absolute Value and Quadratic Functions 9 Solution.. The relation R consists of all points (, ) whose -coordinate is greater than. If we graph =, then we want all of the points in the plane above the points on the graph. Dotting the graph of = as we have done before to indicate that the points on the graph itself are not in the relation, we get the shaded region below on the left.. For a point to be in S, its -coordinate must be less than or equal to the -coordinate on the parabola =. This is the set of all points below or on the parabola =. The graph of R The graph of S. Finall, the relation T takes the points whose -coordinates satisf both the conditions given in R and those of S. Thus we shade the region between = and =, keeping those points on the parabola, but not the points on =. To get an accurate graph, we need to find where these two graphs intersect, so we set =. Proceeding as before, breaking this equation into cases, we get =,. Graphing ields The graph of T

13 0 Linear and Quadratic Functions.. Eercises In Eercises -, solve the inequalit. Write our answer using interval notation > 0. + < < > 7. < 7 8. < < < < > < < >. > < <. The profit, in dollars, made b selling bottles of 00% All-Natural Certified Free-Trade Organic Sasquatch Tonic is given b P () = + 00, for 0. How man bottles of tonic must be sold to make at least $0 in profit?. Suppose C() = 0 + 7, 0 represents the costs, in hundreds of dollars, to produce thousand pens. Find the number of pens which can be produced for no more than $00.. The temperature T, in degrees Fahrenheit, t hours after 6 AM is given b T (t) = t +8t+, for 0 t. When is it warmer than Fahrenheit?

14 . Inequalities with Absolute Value and Quadratic Functions 6. The height h in feet of a model rocket above the ground t seconds after lift-off is given b h(t) = t + 00t, for 0 t 0. When is the rocket at least 0 feet off the ground? Round our answer to two decimal places. 7. If a slingshot is used to shoot a marble straight up into the air from meters above the ground with an initial velocit of 0 meters per second, for what values of time t will the marble be over meters above the ground? (Refer to Eercise in Section. for assistance if needed.) Round our answers to two decimal places. 8. What temperature values in degrees Celsius are equivalent to the temperature range 0 F to 9 F? (Refer to Eercise in Section. for assistance if needed.) In Eercises 9 -, write and solve an inequalit involving absolute values for the given statement. 9. Find all real numbers so that is within units of. 0. Find all real numbers so that is within units of.. Find all real numbers so that is within unit of.. Find all real numbers so that is at least 7 units awa from.. The surface area S of a cube with edge length is given b S() = 6 for > 0. Suppose the cubes our compan manufactures are supposed to have a surface area of eactl square centimeters, but the machines ou own are old and cannot alwas make a cube with the precise surface area desired. Write an inequalit using absolute value that sas the surface area of a given cube is no more than square centimeters awa (high or low) from the target of square centimeters. Solve the inequalit and write our answer using interval notation.. Suppose f is a function, L is a real number and ε is a positive number. Discuss with our classmates what the inequalit f() L < ε means algebraicall and graphicall. 6 In Eercises - 0, sketch the graph of the relation.. R = {(, ) : } 6. R = { (, ) : > + } 7. R = {(, ) : < + } 8. R = { (, ) : < + } 9. R = {(, ) : < < } 0. R = { (, ) : < }. Prove the second, third and fourth parts of Theorem.. 6 Understanding this tpe of inequalit is reall important in Calculus.

15 Linear and Quadratic Functions.. Answers. [, ]. (, ) ( 7 8 7, ). (, ). (, ] [, ). No solution 6. (, ) 7. (, ] [6, ) 8. [, ) (, 6] 9. [ 7, 6 ] 0. (, ) (, ). (, ] [6, ). (, ). No Solution.. [ 7,. (, ) ] 6. (, ) 7. (, ] [, ) 8. (, ) (, ) 9. No solution 0. (, ). {}. No solution. [, ]. (0, ) ( ) ( ) (., 6 + 6, 6. 7., i h 7, 0 0, i h, 7, 7 6 ] [ ) + 7 6, 8. [ 7, + 7 ] [, ] 9. (, ) 0. (, ] {0} [, ). [ 6, ] [, ) ( ). (, ), + 7. P () 0 on [0, ]. This means anwhere between 0 and bottles of tonic need to be sold to earn at least $0 in profit.. C() on [, 8]. This means anwhere between 000 and 8000 pens can be produced and the cost will not eceed $00.. T (t) > on (8, 8 + ) (.7,.6), which corresponds to between 7: AM (.7 hours after 6 AM) to 8:8 PM (.6 hours after 6 AM.) However, since the model is valid onl for t, 0 t, we restrict our answer and find it is warmer than Fahrenheit from 7: AM to 6 PM.

16 . Inequalities with Absolute Value and Quadratic Functions 6. h(t) 0 on [0, 0 + ] [.9, 7.07]. This means the rocket is at least 0 feet off the ground between.9 and 7.07 seconds after lift off. 7. s(t) =.9t + 0t +. s(t) > on (approimatel) (.,.68). This means between. and.68 seconds after it is launched into the air, the marble is more than feet off the ground. 8. From our previous work C(F ) = 9 (F ) so 0 F 9 becomes 0 C. 9., [, 6] 0. +, [,., [, ] [, ] ]. 7, (, ] [, ). Solving S(), and disregarding the negative solutions ields [.0,.79]. The edge length must be within.0 and.79 centimeters. [, ]

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