Solving Inequalities

Simple Inequalities contain one of the following signs < ,> ,\leq ,\: \: or\: \: \geq.

When solving an inequality, one is trying to find all the valid solutions.

Example:                  x+1< 4

                                      x< 3         all real nos. less than 3 

An inequality can be bounded or unbounded / open or closed.  We use rounded brackets to indicate an open ended inequality and we use square brackets to indicate a closed inequality.  If an inequality goes to infinity, either positive or negative infinity, it is considered unbounded.  If there are numbers at the ends of the graph of an inequality it is called bounded.

Examples:            \left ( -3,5 \right ]             -3< x\leq 5             bounded

                             \left ( -3,\infty \right )            x> -3                 unbounded

                             \left [ 0,2 \right ]                0\leq x\leq 2                 bounded

                            \left ( -\infty ,\infty \right )          -\infty < x< \infty        unbounded

How to graph inequalities:

Linear Inequalities \to x^{1}          < \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \leftarrow )

                                                   \geq \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: [ \rightarrow

                                          a< x\leq b \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: (\underset{a}{\leftarrow} \underset{b}{\rightarrow}]        double inequality

Absolute Value Inequalities \rightarrow \left | x^{1} \right |

                                                  \left | \square \right |< \; \; \; \; \; \; \; \; \; \; \; \left ( \rightarrow \leftarrow \right )

                                                  \left | \square \right |> \; \; \; \; \; \; \; \; \; \; \; \leftarrow )\; \; \; \; (\rightarrow

                                                  \left | \square \right |= \; \; \; \; \; \; \; \; \; \; \; \dot{-}\; \; \;\; \; \; \dot{+}

Properties of Inequalities

1.  Transitive            a< b\: \: and \: \: b< c,\: \: \rightarrow \: a< c

2.  Addition of Inequalities

                                  a< b\: \: and \: \: c< d,\: \: \rightarrow \: a+c< b+d

3.  Addition of a Constant

                                  a< b\: \: \rightarrow \: a+c< b+c

4.  Multiplication by a Constant

                                  \left ( + \right )\: \: c> 0,\: \: \: a< b\: \: \rightarrow \: a*c< b*c

                                  \left ( - \right )\: \: c< 0,\: \: \: a< b\: \: \rightarrow \: a*c> b*c

Examples:        a)\: \: 4x< 12

                                       x< 3; \; \; \; \; \; \; (\infty ,3)\; \; \; \; \; \; \; \; \underset{\: \; \; \; \; 3}{\leftarrow})       all reals less than 3

                          b)\: \: \: 2x+7\leq 3+4x     

                                 -2x\leq -4        

                                             x\geq 2\; \; \; \; \; \; \; [2,\infty )\; \; \; \; \; \; \; \; [\underset{2\; \; \; \; \; }{\rightarrow}      all reals greater than or equal to 2

                          c)\: \: \: \frac{1}{2}\left ( 8x+1 \right )\geq 3x+\frac{5}{2}     

                                   4x+\frac{1}{2}\geq 3x+\frac{5}{2}      

                                          x\geq \frac{4}{2}

                                                   x\geq 2\; \; \; \; \; \; \; [2,\infty )\; \; \; \; \; \; [\underset{2\; \; \; \; }{\rightarrow}       all reals greater than or equal to 2

                                 d)\; \; 1< 2x+3< 9\; \; \; \; \rightarrow \: \: 1< 2x+3\: \: and \; \; 2x+3<9

                                                                            -2< 2x\: \: \; \; \; \; \; \; \; \; \; \; \; 2x<6

                                                                             -1< x\: \: \: \: \; \; \; \; \; \; \; \; \; \; \; x<3

                                       -1< x< 3\; \; \; \; \; \; \; \left ( -1,3 \right )\; \; \; \; \; \; \; \left ( \underset{-1\: \: \: \: \: }{\leftarrow}\underset{\: \: \: \: \: 3}{\rightarrow} \right )    all reals between -1 and 3

                         e)\; \; \; \left | \frac{x-3}{2} \right |\geq 5\; \; \; \; \; \; think \; \; \; \underset{\; \; -5}{\leftarrow}\: \: \; \; \; \: \: \underset{5\; \; \; \; }{\rightarrow}             

                                               \frac{x-3}{2}\leq -5                 \frac{x-3}{2}\geq 5

                                              x-3\leq -10                 x-3\geq 10

                                                    x\leq -7                     x\geq 13

      all reals such that x is less than or equal to -7 and x is greater than or equal to 13

\underset{\; \; \; -7}{\leftarrow}]\; \; \; \; \; [\underset{13\; \; \; \; }{\rightarrow}

                    f)\; \; \; \left | 9-2x \right |-2< -1

                              \left | 9-2x \right |< 1\; \; \; think\; \; \; \left ( \underset{-1\; \; \; \; \; \; \; }{\rightarrow}\underset{\: \: \: \: \: \: \: \: 1}{\leftarrow}\right )

                                       9-2x> -1                 9-2x< 1

                                          -2x> -10                   -2x< -8

                                              x< 5                             x> 4

                                     4< x< 5\; \; \; \;\; \; \; \; \; \; \left ( \underset{4\: \; \; \; }{\rightarrow}\underset{\; \; \; \; \; 5 }{\leftarrow} \right )    all reals between 4 and 5

Other types of Inequalities:

Polynomial Inequalities:

Find its ‘zeros’.  Zeros in order divide a number line into intervals.  Zeros are called critical numbers and the intervals are called test intervals.

A quadratic inequality means the degree of x is 2.  When the degree is 2, there are 2 zeros.  Just like absolute value inequalities, the solution for an inequality of degree 2 is a bounded interval between the two zeros if the inequality is less than and/or equal to a positive constant.  If the inequality is greater than and/or equal to a positive constant, the solution consists of two intervals.  One interval starts at negative infinity and goes to the lower bound and the other starts at the upper bound and goes to positive infinity.

             x^{2}\rightarrow 2 \: \: zeros            < \: \: \rightarrow \; \; \; \; \left ( \rightarrow \leftarrow \right )                > \; \; \rightarrow \; \; \left \{ \leftarrow )\; \; \; (\rightarrow \right \}

Examples:         a)\; \; \; x^{2}\leq 9

                      x^{2}=9                 \rightarrow x=\pm 3             \rightarrow Zeros = -3 \; \; and \: \; 3

    The test intervals are:             (-\infty ,-3)\; \; \; \; \; \; \; \left ( -3,3 \right )\; \; \; \; \; \; \; \left ( 3,\infty \right )   

                          Test:    x=-4\; \; \rightarrow 16\nleqslant 9\; \; \; \; \; x=1\; \; \rightarrow 1\leq 9\; \; \; \; x=5\; \; \; 25\nleqslant 9

     The solution to a) is -3\leq x\leq 3\; \; \; \; \; \; \; \left [ \underset{-3\; \; \; \; }{\rightarrow} \underset{\; \; \; \; 3}{\leftarrow}\right ]         all real numbers between -3 and 3

                   b)\; \; \; x^{2}+4x+4\geq 9\; \; \; \; \; \; think \; \; \; \leftarrow ]\; \; \; [\rightarrow

                                        \left ( x+2 \right )^{2}=9            zeros   \rightarrow x+2=\pm 3

                                                                                            x=-2\pm 3

                                                                                            x=-5,1

                  all reals   x\leq -5  and   x\geq 1             \underset{\; \; \; -5}{\leftarrow}]\; \; \; and\; \; [\underset{\; \; 1}{\rightarrow}

            c)\; \; \; x^{3}-2x^{2}-9x-2\geq -20            Think:  find the zeros and test intervals

                 x^{3}-2x^{2}-9x+18=0

                 x\left ( x^{2}-9 \right )-2\left ( x^{2}-9 \right )=0

                 \left ( x^{2} -9\right )\left ( x-2 \right )=0

                 \left ( x+3 \right )\left ( x-3 \right )\left ( x-2 \right )=0

                  Zeros      \rightarrow \; -3,3,2        in order   \rightarrow \; -3,2,3    

                                       test intervals   (-\infty ,-3],\left [ -3,2 \right ],[2,3],[3,\infty )

                            (-\infty ,-3]            x=-4\; \; \rightarrow \left ( -4 \right )^{3}-2\left ( -4 \right )^{2}-9\left ( -4 \right )-2\geq -20

                                                                               -64-32+36-2=-62\ngeqslant -20

                            \left [ -3,2 \right ]                  x=-2\; \; \rightarrow \left ( -2 \right )^{3}-2\left ( -2 \right )^{2}-9\left ( -2 \right )-2\geq -20  

                                                                                        -8-8+18-2=0\geq -20                     

                             \left [ 2,3 \right ]                    x=2.5\; \; \rightarrow \left ( 2.5 \right )^{3}-2\left ( 2.5 \right )^{2}-9\left ( 2.5 \right )-2\geq -20    

                                                                15.625-12.5-22.5-2=-21,375\ngeqslant -20   

                             \left [ 3,\infty \right )                   x=4\; \; \rightarrow \left ( 4\right )^{3}-2\left ( 4 \right )^{2}-9\left ( 4\right )-2\geq -20   

                                                                                    64-32-36-2=-6\geq -20       

            all real such that -3\leq x\leq 2  and x\geq 3

Rational Inequalities have restrictions.  Any value of x that makes the denominator equal to 0 is a restriction.  Any value of x that makes the value under a square root sign, negative, is also a restriction.  The critical numbers that divide the number line into test intervals includes the zeros and the restrictions. 

Examples:       a)\; \; \; \frac{x+6}{x+1}< 2         Find the R [restrictions] and the Zeros.

                                 \frac{x+6}{x+1}-2=0            R = -1 [ -1 makes the D = 0]

                                 x+6-2\left ( x+1 \right )=0         LCD is (x+1)

                                 x+6-2x-2=0

                                         -x+4=0

                             -x=-4\; \; \rightarrow x=4             Zero = 4

                         Critical Numbers [ Zeros plus R ] = -1, 4

                         Test Intervals =  \left ( -\infty ,-1 \right ),\; \left ( -1,4 \right ),\; \left ( 4,\infty \right )

                         \left ( -\infty ,-1 \right )            x=-2\; \; \; \rightarrow \frac{-2+6}{-2+1}=\frac{4}{-1}=-4< 2

                         \left ( -1,4 \right )                 x=1\; \; \; \rightarrow \frac{1+6}{1+1}=\frac{7}{2}=3.5\nless 2

                         \left ( 4,\infty \right )                   x=5\; \; \; \rightarrow \frac{5+6}{5+1}=\frac{11}{6}=1\frac{5}{6}< 2

             all reals \left \{ x< -1\; \; and\; x> 4 \right \} .

                 b)\; \; \frac{x^{2}+2x}{x^{2}-9}\leq 0             Find the critical numbers.

                     Zeros   \rightarrow x\left ( x+2\right )=0                      R   \rightarrow \pm 3

                                      x=0,-2

                     Critical numbers in order  \rightarrow -3,-2,0,3

                      Test Intervals   \rightarrow \left ( -\infty ,-3 \right ),\; \; \; \left ( -3,-2 \right ],\; \; \left [ -2,0 \right ],\; \; \left [ 0,3 \right ),\; \; \left ( 3,\infty \right )

                      \left ( -\infty ,-3 \right )           x=-4\; \; \rightarrow \frac{16-4}{16-9}=\frac{12}{7}\nleqslant 0

                      \left ( -3,-2 \right ]              x=-2.5\; \; \rightarrow \frac{6.25-4}{6.25-9}=\frac{2.25}{-2.75}=-0.\bar{81}\leq 0

                      \left [ -2,0 \right ]                 x=-1\; \; \rightarrow \frac{1-2}{1-9}=\frac{-1}{-8}=\frac{1}{8}\nleqslant 0

                      \left [0,3 \right )                   x=1\; \; \rightarrow \frac{1+2}{1-9}=\frac{3}{-8}=-\frac{3}{8}\leq 0

                      \left (3,\infty \right )                x=4\; \; \rightarrow \frac{16+8}{16-9}=\frac{24}{7}\nleqslant 0

          all reals such that   -3< x\leq -2   and   0\leq x< 3

                   c)\; \; \sqrt{\frac{x}{x^{2}-2x-35}}\geq 0

                     \frac{x}{\left ( x-7 \right )\left ( x+5 \right )}=0            Zeros = 0,   R = 7, -5

                      Critical Numbers in order:     -5, 0, 7

                      Test Intervals:       \; \; \rightarrow \left ( -\infty ,-5 \right ),\; \; \left ( -5,0 \right ],\; \; \left [ 0,7 \right ),\; \; \left ( 7,\infty \right )     

                       \left ( -\infty ,-5 \right )            x=-6\; \; \rightarrow \sqrt{\frac{-6}{36+12-35}}=\sqrt{-\frac{6}{13}}\rightarrow no \: \; solution

                       \left ( -5,0 \right ]                 x=-1\; \; \rightarrow \sqrt{\frac{-1}{1+2-35}}=\sqrt{\frac{1}{33}}\geq 0

                       \left [ 0,7 \right )                    x=1\; \; \rightarrow \sqrt{\frac{1}{1+2-35}}=\sqrt{-\frac{1}{33}}\; \; \; no\: \; \; solution

                       \left ( 7,\infty \right )                  x=8\; \; \rightarrow \sqrt{\frac{8}{64-16-35}}=\sqrt{\frac{8}{13}}\geq 0

             all reals such that  -5< x\leq 0  and   x> 7 .

 

Filed under: Pre-requisites for Pre-Calculus

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