2.4 Inequalities. 154 Linear and Quadratic Functions

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1 Linear and Quadratic Functions. Inequalities In this section, not onl do we develop techniques for solving various classes of inequalities analticall, we also look at them graphicall. The net 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, ). 8 = g() = f() To see the connection between the graph and the algebra, we recall the Fundamental Graphing Principle for Functions in Section.: 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 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. We sa 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. 8 8 = f() = g() = g() = f() f() < g() f() > g() 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 precisel the values where the graphs of = f() and = g() intersect. The solutions to f() < g() are precisel the values where the graph of = f() is below the graph of = g(). The solutions to f() > g() are precisel the 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 Linear and Quadratic Functions Eample... The graphs of f and g are below. The graph of = f() resembles the upside down shape of an absolute value function while the graph of = g() resembles a parabola. 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 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 has no solution. For c 0, > c is equivalent to < c or > c. For c < 0, > c is 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 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 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.. To solve, we seek solutions to > as well as solutions to =. From Theorem., > is equivalent to < or >. From Theorem., = is equivalent to = or =. Combining these equations with the inequalities, we solve or. Our answer is or, which, in interval notation is (, ] [, ). Graphicall, we have

5 8 Linear and Quadratic Functions We see the graph of = (the ) is above the horizontal line = for < and >, and, hence, this is where >. The two graphs intersect when = and =, 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 + > or + <. Rewriting, we now have < + < so that < <. In interval notation, we write (, ). Graphicall we see 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 > and so < or >. We get < or >. Our solution to the first inequalit is then (, ) (, ). For, we combine results in Theorems. and. to get so that, or [, ]. Our solution to < is comprised of values of which satisf both parts of the inequalit, and so we take what s called the set theoretic intersection of (, ) (, ) with [, ] to obtain [, ) (, ]. Graphicall, we see the graph of = is between the horizontal lines = and = for values between and as well as those between and. Including the values where = and = intersect, we get

6 . Inequalities We need to eercise some special caution when solving + +. 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, 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 < 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() = and g() = 0, then solving < 0 corresponds graphicall to finding the values of for which the graph of = f() = (the parabola) is below the graph of = g() = 0 (the -ais.) We ve provided the graph again for reference.

7 0 Linear and Quadratic Functions = 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 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() = (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 or final answer is [, ]. To check 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 two graphs intersect (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 Linear and Quadratic Functions (+) 0 ( ) 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. 8 (+) 0 ( ) 0 (+) + 0. To solve 9 +, as before, we solve Setting f() = 9 + = 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 9 + 0, we are looking for

10 . Inequalities values where 9 + < 0 as well as where 9 + = 0. Looking at our sign diagram, there are no places where 9 + < 0 (there are no ( )), so our solution is onl = (where 9 + = 0). We write this as { }. Graphicall, we solve 9 + b graphing g() = 9 + and h() =. We are looking for the values where the graph of g is below the graph of h (for 9 + < ) and where the two graphs intersect (9 + = ). We see the line and the parabola touch at (, 8), but the parabola is alwas above the line otherwise. (+) 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 ) =. Solving choose = and = as our test values. We find f( ) = and f ( 0 for < gives us [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() =. To solve g() 0, we have [, ). (+) 0 ( ) 0 ( ) 0 (+) Combining these into one sign diagram, we get our solution is [0, ]. Graphicall, to check, we set h() = and i() = and look for the values In this case, we sa the line = is tangent to = 9 + at (, 8). Finding tangent lines to arbitrar functions is the stuff of legends, I mean, Calculus.

11 Linear and Quadratic Functions 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 It is quite possible to encounter inequalities where the analtical methods developed so far will fail us. In this case, we resort to using the graphing calculator to approimate the solution, as the net eample illustrates. Eample... Suppose the revenue R, in thousands of dollars, from producing and selling hundred LCD TVs is given b R() = + + for 0, while the cost, in thousands of dollars, to produce hundred LCD TVs is given b C() = 00 + for 0. How man TVs, to the nearest TV, should be produced to make a profit? Solution. Recall that profit = revenue cost. If we let P denote the profit, in thousands of dollars, which results from producing and selling hundred TVs then P () = R() C() = ( + + ) (00 + ) = +, where 0. If we want to make a profit, then we need to solve P () > 0; in other words, + > 0. We have et to discuss how to go about finding the zeros of P, let alone making a sign diagram for such an animal, as such we resort to the graphing calculator. After finding a suitable window, we get We are looking for the values for which P () > 0, that is, where the graph of P is above the -ais. We make use of the Zero command and find two -intercepts. The procedure, as we shall see in Chapter is identical to what we have developed here.

12 . Inequalities We remember that denotes the number of TVs in hundreds, so if we are to find our solution using the calculator, we need our answer to two decimal places. The zero.... corresponds to.... TVs. Since we can t make a fractional part of a TV, we round this up to TVs. The other zero seems dead on at, which corresponds to 00 TVs. Hence to make a profit, we should produce (and sell) between and 99 TVs, inclusive. 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... Sketch the following relations.. R = {(, ) : > }.. S = {(, ) : }.. T = {(, ) : < }. 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 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 Note the -coordinates of the points here aren t registered as 0. The are epressed in Scientific Notation. For instance, E corresponds to , which is prett close in the calculator s ees to 0. but not a Mathematician s Notice that P () < 0 and P () > 0 so we need to round up to in order to make a profit.

13 Linear and Quadratic Functions. Finall, the relation T takes the points whose -coordinates satisf both the conditions in R and S. So 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