Exercises - Solving Inequalities

  1. Solve the following:

    1. $\displaystyle{3 \left[ \frac{2}{3} (2x-1) - (x+4) \right] \lt -(5 - 3x) + \frac{3}{2} (4x-2)}$   $\ans{\displaystyle{x \gt -\frac{3}{4}}}$

    2. $\displaystyle{3 - 4(x+1) \le x - 3 \left[ \frac{1}{2} (2x-3) \right]}$   $\ans{\displaystyle{x \ge - \frac{11}{4}}}$

  2. Solve the following inequalities

    1. $\displaystyle{x^2-2x-3 \lt 0}$   $\ans{\displaystyle{(-1,3)}}$

    2. $\displaystyle{-x^2-2x+3 \gt 0}$   $\ans{\displaystyle{(-3,1)}}$

    3. $\displaystyle{-x^2-2x+3 \le 0}$   $\ans{\displaystyle{(-\infty,-3] \cup [1, \infty)}}$

    4. $\displaystyle{x^2+2x \le 3}$   $\ans{\displaystyle{[-3,1]}}$

    5. $\displaystyle{x^2 \le 2x + 3}$   $\ans{\displaystyle{[-1,3]}}$

    6. $\displaystyle{x^3-6x^2 \le 8-12x}$   $\ans{\displaystyle{(-\infty,2]}}$

    7. $\displaystyle{x^3-x^2 \gt 3-3x}$   $\ans{\displaystyle{(1,\infty)}}$

  3. Determine which values of $x$ make the given expression positive, negative, and zero.

    1. $\displaystyle{-2(x-5)^2}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos: never}\\ \textrm{neg: } (-\infty,5) \cup (5,\infty)\\ \textrm{zero: }x = 5 \end{array} }}$

    2. $\displaystyle{x^2-3x-40}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos:} (-\infty,-5) \cub (8,\infty)\\ \textrm{neg: (-5,8)} \\ \textrm{zero: x=-5,8} \end{array} }}$

    3. $\displaystyle{x^3+3x^2+3x+1}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos:} (-1,\infty)\\ \textrm{neg: } (-\infty, -1)\\ \textrm{zero: x = -1} \end{array} }}$

    4. $\displaystyle{x^2-7x+1}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos:} \displaystyle{ \left( -\infty,\frac{7-3\sqrt{5}}{2} \right) \cup \left( \frac{7 + 3\sqrt{5}}{2},\infty \right) }\\ \textrm{neg: } \displaystyle{ \left( \frac{7-3\sqrt{5}}{2}, \frac{7+3\sqrt{5}}{2} \right)}\\ \textrm{zero: } \displaystyle{ x = \frac{7 \pm 3\sqrt{5}}{2}} \end{array} }}$

    5. $\displaystyle{(x^2-4x+4) - (4x^3-2x^2+12x-8)}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos:} (-\infty, 3/4)\\ \textrm{neg: } (3/4, \infty)\\ \textrm{zero: x = 3/4} \end{array} }}$

    6. $\displaystyle{(x+1) (x-2)^2 (x+3) (x-5)^7}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos:} (-3,-1) \cup (5,\infty)\\ \textrm{neg: } (-\infty,-3) \cup (-1,2) \cup (2,5)\\ \textrm{zero: } x = -3,-1,2, \textrm{ or } 5 \end{array} }}$

    7. $\displaystyle{x^{2/3}(x-3)^{3/5}}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos:} (3,\infty)\\ \textrm{neg: } (-\infty,0) \cup (0,3)\\ \textrm{zero: } x = 0, 3 \end{array} }}$

    8. $\displaystyle{(x-1)^{1/3}(x-2)^{2/3} + (x-1)^{4/3}(x-2)^{5/3}}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos:} (1,2) \cup (2,\infty)\\ \textrm{neg: } (-\infty,1)\\ \textrm{zero: } x = 1,2 \end{array} }}$

    9. $\displaystyle{\frac{(x+3)^2(5-x)^{\frac{3}{5}}}{(x^2-4)(x^2+x+1)}}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos:} (-\infty,-3) \cup (-3,-2) \cup (2,5)\\ \textrm{neg: } (-2,2) \cup (5,\infty)\\ \textrm{zero: } x = -3, \textrm{ or }5 \end{array} }}$

    10. $\displaystyle{\frac{(x-2)^3(x^2-9)^2}{x(x^2+6x+9)}}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{pos:} (-\infty,-3) \cup (-3,0) \cup (2,3) \cup (3,\infty)\\ \textrm{neg: } (0,2)\\ \textrm{zero: } x = 2, \textrm{ or } 3 \end{array} }}$

  4. Solve the following:

    1. $\displaystyle{\left| 5-3x \right| \gt 4}$   $\ans{\displaystyle{(-\infty,\frac{1}{3}) \cup (3,\infty)}}$

    2. $\displaystyle{\left| 2 - 5x \right| \lt 3}$   $\ans{\displaystyle{(-\frac{1}{5},1)}}$

    3. $\displaystyle{\left| 7-2x \right| \lt 5}$   $\ans{\displaystyle{(1,6)}}$

    4. $\displaystyle{\left| \frac{4x-1}{2} \right| \ge 3}$   $\ans{\displaystyle{(-\infty,-\frac{5}{4}] \cup [ \frac{7}{4},\infty)}}$

    5. $\displaystyle{\left| 3x+4 \right| = 9}$   $\ans{\displaystyle{x=\frac{5}{3},-\frac{13}{3}}}$

  5. Solve $\displaystyle{\frac{(x-2)^{1/3}(2x+3)^2}{(x+5)^3(x^2+4)} \ge 0}$   $\ans{\displaystyle{(-\infty,-5) \cup \{-3/2\} \cup [2,\infty)}}$

  6. For what values of $x$ do the following expressions represent real values?

    1. $\displaystyle{\sqrt{x^2-5}}$   $\ans{\displaystyle{(-\infty,-5] \cup [5,\infty)}}$

    2. $\displaystyle{\sqrt{\frac{x-4}{x+4}}}$   $\ans{\displaystyle{(-\infty,-4) \cup [4,\infty)}}$