1) Which form is best when asked to determine the values of x for which the expression is undefined, equal to zero, positive or negative? Why?

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1 Solving Inequalities Prepared by: Sa diyya Hendrickson Name: Date: Consider the expanded and simplified forms of the following expression: Expanded Form: x + x 3x Simplified Form: x(x + 1) (3x ) 1) Which form is best when asked to determine the values of x for which the expression is undefined, equal to zero, positive or negative? Why? ) Determine the values of x for which the expression above is (i) undefined, (ii) equal to zero, (iii) positive or (iv) negative. Making Math Possible 1 of 10 c Sa diyya Hendrickson

2 Graphical Approach Consider the inequality f(x) < g(x) where f(x) = x + 1 and g(x) = 3x 5. Here, we are being asked to determine all x-values such that the value of f(x) is less than the value of g(x), or equivalently, such that the graph of f(x) is below the graph of g(x). Next, let s consider an inequality involving at least one nonlinear function. Let f(x) = 1 x such that f(x) < g(x). This results in the following equivalent inequalities: and g(x) =x, I 1 : 1 x <x () I : 1 x x<0 To get from I 1 to I, we simply subtracted x from both sides (by properties of inequalities). 1. I 1 asks us to determine all values of x such that f(x) = 1 x is below g(x) =x.. I asks us to consider the function h(x) = 1 x x and determine where h(x) is negative (i.e. where the graph of h(x) falls below the x-axis). By getting a zero on one side, we dramatically simplify the exercise. This is because we only have to compare the function with y = 0 (i.e. the x-axis), to determine where it s positive, negative and equal to zero! Making Math Possible of 10 c Sa diyya Hendrickson

3 Graphical Approach Proceeding with I, we want to determine where h(x) = 1 x x is strictly negative. From the graph, we see that x ( 1, 0) [ (1, 1) Given that we will be learning to solve various kinds of inequalities, a fair question to ask is: When does a function usually change sign? Making Math Possible 3 of 10 c Sa diyya Hendrickson

4 Graphs to Algebra Below we will summarize some of the important lessons we ve learned thus far. I. Linear Inequalities A linear inequality only involves linear terms. The process for solving a linear inequality is almost identical to solving a linear equation. Our ultimate goal is still to isolate the variable! However, one rule for inequalities requires that we proceed with caution as we are doing the opposite to both sides! Recall that when multiplying/dividing both sides of the inequality by a negative number, we must reverse the direction of the inequality symbol! In other words: If c<0, then a apple b () ac bc and a apple b () a b c c II. Nonlinear Inequalities A nonlinear inequality involves at least one nonlinear term. The graphs on the previous page were helpful because they provided us with visual aids. However, graphing functions to solve inequalities is not a rigorous approach. It is also not the most practical. Below we will discuss how to use our findings to develop an approach that is purely algebraic! Making Math Possible 4 of 10 c Sa diyya Hendrickson

5 Linear Inequalities Exercise L1: Solve the inequality x 5 apple 1 3x Solution: Isolate x by doing the opposite (just as you would for linear equations), while making sure to reverse the direction of the inequality whenever you multiply or divide both sides by a negative number! x 5 apple 1 3x x 5 +3x apple 1 3x +3x add 3x to both sides 5x 5 apple1 collect like terms 5x 5 +5apple1 +5 add 5 to both sides 5x apple 6 5x 5 apple 6 5 x apple 6 5 collect like terms divide both sides by 5 simplify Therefore, the solution is given by: x ( 1, 6/5] Exercise L: Solve the pair of simultaneous inequalities < 3 4x apple 7 Solution: For this exercise, we have two approaches: 1 Solve the following inequalities separately: < 3 4x and 3 4x apple 7. Then, take the intersection of the two solution sets. Isolate x by doing the opposite across both inequalities at the same time. We will proceed with the second approach, because it s the most e cient! < 3 4x apple 7 3 < 3 4x 3 apple 7 3 subtract 3 across both inequalities 1 < 4x apple 4 collect like terms 1 4 > 4x divide by 4 and reverse the inequalities 1 >x 1 simplify 4 From the final line, we have 1 apple x< 1 4. Therefore, our solution is given by: x [ 1, 1/4) Making Math Possible 5 of 10 c Sa diyya Hendrickson

6 Nonlinear Inequalities Consider the following example of a nonlinear inequality: x +x apple x 3 +x Before we discuss strategies for solving an inequality of this kind, we will make an important observation: Many students begin this exercise by cross-multiplying, as they would with equations: x +x apple x 3 +x (x 3 +x ) apple (x +x) Why is this incorrect? Notice that the expressions x +x and x 3 +x are not positive for all values of x. Take a moment to verify that when x = 1, x +x = 1 and that when x = 3, x 3 +x = 9. Therefore, we cannot use this approach until we first determine all of the cases for which these expressions are positive and negative. Then, we must solve each case separately. If an expression is negative in a particular case, we must reverse the direction of the inequality. Even though you can fix the error by taking cases, this approach is not recommended because it leaves a lot of room for error and the solutions tend to be poorly organized and presented. Below we will discuss an approach that is safe, organized and e cient! Exercise NL1: Solve the nonlinear inequality x +x apple x 3 +x S1 Get a zero on one side, using addition and/or subtraction. Why? So that we won t have to compare two expressions. We will simply have to determine when the new expression on the LS (or RS) is greater than zero (i.e. positive), less than zero (i.e. negative) and/or equal to zero. x +x x 3 +x apple 0 S Simplify (without reducing) the new expression on the nonzero side. i.e. Attempt to factor, get an LCD (if necessary), collect like terms, and try to factor again (if necessary)! On the LS we have a rational expression. So, we will begin by finding an LCD. D 1 = x +x = x(x + ) D = x 3 +x = x (x + ) ) LCD = x (x + ) Making Math Possible 6 of 10 c Sa diyya Hendrickson

7 Nonlinear Inequalities Exercise NL1 (Continued) Making the appropriate adjustments, we have: x +x x 3 +x apple 0 LCD ) LCD ) ) (x) (x +x)(x) x x (x + ) apple 0 (x 1) x (x + ) apple 0 x 3 +x apple 0 S3 Determine the x-values that give: 1 Restricted values (i.e. where the expression is undefined) x-intercepts (i.e. where the expression is equal to zero) Here, we are finding all possible values where the expression may be changing sign (from positive to negative or vice versa). 1 Restrictions: DENOM 6= 0 ) x (x + ) 6= 0 ZPP ) x 6= 0 and x +6= 0 ) x 6= 0 and x 6= (Note: never shade on number line) Equal to zero! NUM = 0 and DENOM 6= 0 ) x 1=0 ) x =1 We shade this value because the exercise asks for apple 0 S4 Use a number line/chart to test values and determine appropriate intervals. In the chart below, we are looking for negative signs because we require the expression to be less than or equal to zero. Therefore, the solution is given by: x (, 0) [ (0, 1]. Making Math Possible 7 of 10 c Sa diyya Hendrickson

8 Absolute Value Inequalities Note that there are di erent types of absolute value inequalities. Below we will discuss two, both of which have only one term involving absolute value. Type I One absolute value term and all other terms are constants. Exercise AV1: Solve the absolute value inequality 3x 5 3 > 1 S1 Isolate the absolute value term by moving all constants/numbers to the other side (if necessary). i.e. Create an inequality of the form: a >b. 3x 5 3 > 1 () 3x 5 > 4 S Use the properties of absolute value inequalities to produce the relevant inequalities. Recall that the properties are as follows: Inequality Equivalent form Line Graph 1. x <c c<x<c. x applec c apple x apple c 3. x >c x< c or x>c 4. x c x apple c or x c Here, we have 3x 5 > 4, which means that a = 3x 5 is more than 4 units away from zero! Thus, we have the following two cases to consider: Case 1: 3x 5 < 4 Case : 3x 5 > 4 S3 Solve the inequality under each case. Case 1 Case 3x 5 < 4 3x 5 > 4 3x 5 +5 < x 5 +5 > x < 1 3x > 9 3x 3 < 1 3 3x 3 > 9 3 ) x< 1 3 or x>3 Therefore, our solution is given by: x ( 1, 1/3) [ (3, 1) Making Math Possible 8 of 10 c Sa diyya Hendrickson

9 Absolute Value Inequalities Type II One absolute value term along with at least one non-constant term (i.e. a term involving the variable to some non-zero power). Consider the following example: x 3 x + 5. How is this inequality di erent from the previous Type I absolute value inequality? 1 Firstly, the distance of x 3 from 0 is greater than or equal to x + 5, which changes as x changes. Second, because x + 5 is not constant, we cannot use the properties of absolute value inequalities as we did in the previous exercise. These di erences require us to proceed with caution. A popular and safe approach is to take cases as dictated by the algebraic definition of absolute value. While doing so, we must make note of the restrictions and only take the values satisfying the restrictions of each case! Recall that: 8 >< a if a 0 (Case 1) a = >: a if a<0 (Case ) Exercise AV: Solve the equation x 3 x + 5. S1 Get rid of the absolute value bars using the definition of absolute value. By definition, we have: 8 >< x 3 = >: x 3 if x 3 0 (Case 1) (x 3) if x 3 < 0 (Case ) S Solve the linear/nonlinear inequality under each case, making sure to take the intersection with the restrictions of each case. Considering each case seperately, we have: (i) Case 1: x 3 0 ) x 3 Let s denote our restriction by R 1 =[3, 1) In the interval R 1,wehave: x 3 = x 3 Then, the inequality becomes: x 3 x + 5 Solving, we have: x 3 x x + 5 x Let S 1 =( 1, 8] x 3 5 x x 8 ( 1)( x) apple ( 1)(8) x apple 8 For the Case 1 solution, we must find the following intersection: T 1 = R 1 \ S 1 = ; (the empty set) Making Math Possible 9 of 10 c Sa diyya Hendrickson

10 Absolute Value Inequalities Exercise AV (Continued) (ii) Case : x 3 < 0 ) x<3 Let s denote our restriction by R =[ 1, 3) In the interval R,wehave: x 3 = (x 3) Then, the inequality becomes: (x 3) x + 5 Solving, we have: x + 3 x x + 5 x 3x x x 3x 3 apple 3 Let S =( 1, /3] x apple 3 For the Case solution, we must find the following intersection: T = R \ S =( 1, /3] S3 Take the union of the solutions from Case 1 and Case to find the solution. The final solution is given by: T 1 [ T = ;[( 1, /3] = ( 1, /3] Making Math Possible 10 of 10 c Sa diyya Hendrickson