Solving inequalities. Jackie Nicholas Jacquie Hargreaves Janet Hunter

Transcription

1 Mathematics Learning Centre Solving inequalities Jackie Nicholas Jacquie Hargreaves Janet Hunter c 6 Universit of Sdne

2 Mathematics Learning Centre, Universit of Sdne Solving inequalities In these nots we will look at solving inequalities using both algebraic and graphical techniques. Sometimes it is easier to use an algebraic method and sometimes a graphical one. For the following eamples we will use both, as this allows us to make the connections between the algebra and the graphs. Algebraic method Graphical method. Solve 3. This is a linear inequalit. Remember to reverse the inequalit sign when multipling or dividing b a negative number. 3 (,) 3 When is the line =3 above or on the horizontal line =? From the graph, we see that this is true for.. Solve +3<. This is a quadratic inequalit. Factorise and use a number line. +3 < ( 3)( ) < The critical values are and 3, which divide the number line into three intervals. We take points in each interval to determine the sign of the inequalit; eg use =, = and = as test values. positive negative positive Let = When does the parabola have negative -values? OR When is the parabola under the -ais? From the graph, we see that this happens when <<3. 3 Thus, the solution is <<3.

3 Mathematics Learning Centre, Universit of Sdne 3. Solve. There is a variable in the denominator. Remember that a denominator can never be zero, so in this case. First multipl b the square of the denominator Let =. (,) ( ), ( )( ) 6 Mark the critical values on the number line and test =, =. and =6. positive neg positive 3 6 = is not defined for =. It is a hperbola with vertical asmptote at =. To solve our inequalit we need to find the values of for which the hperbola lies on or under the line =. (, ) is the point of intersection. So, from the graph we see that when <or. Therefore, <or.. Solve 3 <. Consider 3=,. Multipl b we get 3 = 3 = ( )( +) = Therefore, the critical values are, and which divide the number line into four intervals. We can use = 3, =, = and = 6 as test values in the inequalit. The points = 3 and = satisf the inequalit, so the solution is < or<<. (Notice that we had to include as one of our critical values.) Sketch = 3 and then =. Note that second of these functions is not defined for = (, ) 6 (, 3) (,) For what values of does the line lie under the hperbola? From the graph, we see that this happens when < or <<.

4 Mathematics Learning Centre, Universit of Sdne 3 Eample Sketch the graph of = 6. Hence, where possible, a. Solve i. 6 = ii. 6 > iii. 6 = +3 iv. 6 <+3 v. 6 = 3 b. Determine the values of k for which 6 = + k has eactl two solutions. Solution 6 for 3 f() = 6 = ( 6) for <3 = 6. = (9,). (,) -. = (.,3) = a. i. Mark in the graph of =. It is parallel to one arm of the absolute value graph. It has one point of intersection with = 6 = +6(<3) at =.. ii. When is the absolute value graph above the line =? From the graph, when <..

5 Mathematics Learning Centre, Universit of Sdne iii. = + 3 intersects = 6 twice. To solve 6 = + 3, take 6 = 6= + 3 when 3. This gives us the solution = 9. Then take 6 = +6= + 3 when <3 which gives us the solution =. iv. When is the absolute value graph below the line = +3? From the graph, <<9. v. = 3 intersects the absolute value graph at = 3 onl. b. k represents the -intercept of the line = + k. When k = 3, there is one point of intersection. (See (a) (v) above). For k> 3, lines of the form = + k will have two points of intersection. Hence 6 = + k will have two solutions for k> 3.. Eercises. Solve a. p b. p+3 c. 7 9 >. a. Sketch the graph of =( 3). b. Hence solve ( 3). 3. a. Find the points of intersection of the graphs = and =. b. On the same set of aes, sketch the graphs of = and =. c. Using part (ii), or otherwise, write down all the values of for which >. a. Sketch the graph of =. b. Solve <. c. Suppose <a<band consider the points A(a, a ) and B(b, b ) on the graph of =. Find the coordinates of the midpoint M of the segment AB. Eplain wh a + b > a+b. a. Sketch the graphs of = and = on the same diagram. b. Solve >. c. For what values of m does m = have eactl i. two solutions ii. no solutions 6. Solve

6 Mathematics Learning Centre, Universit of Sdne. Solutions. a. b. 3 <p c. < or 3 <<3or>. a. The graph of =( 3) is given below b. From the graph we see that ( 3) when a. The graphs = and = intersect at the points (, ) and (, ). b. The graphs of = and = c. The inequalit is satisfied for < or <<.. a. The graph of =.

7 Mathematics Learning Centre, Universit of Sdne 6 b. < when <. c. The midpoint M of the segment AB has coordinates ( a+b, a + b ). Since the function = is concave up, the -coordinate of M is greater than f( a+b). So, a + b > a+b B M A. a. (.,.) b. >for all <.. c. i. m = has eactl two solutions when <m<. ii. m = has no solutions when <m<. 6. 3