Solving Inequalities Graphically (DP IB Analysis & Approaches (AA): HL): Revision Note

Solving inequalities graphically

How do I solve inequalities graphically?

• Consider the inequality $f\left(x\right)\le g\left(x\right)$, where $f\left(x\right)$ and $g\left(x\right)$ are functions of $x$

  • Subtract $g\left(x\right)$ from both sides to get zero on the right-hand side

    • $f\left(x\right)-g\left(x\right)\le 0$

• Solve the equation $f\left(x\right)-g\left(x\right)=0$ to find the $x$-intercepts of the graph $y=f\left(x\right)-g\left(x\right)$

• Sketch the graph of $y=f\left(x\right)-g\left(x\right)$

  • Use your GDC to help

  • Label its $x$-intercepts

  • The solutions are the range(s) of values of $x$ for which the curve is

    • below the $x$-axis

    • as $f\left(x\right)-g\left(x\right)\le 0$

  • Present your solutions as inequalities

    • e.g. $2\le x\le 10$

• If the inequality in the question had been reversed, $f\left(x\right)\ge g\left(x\right)$

  • then $f\left(x\right)-g\left(x\right)\ge 0$

    • solutions are the values of $x$ where the curve is above the $x$-axis

When do I flip the inequality sign?

• Remember to flip the sign of the inequality when you multiply or divide both sides by a negative number

  • e.g. $-x^2<5-x$ becomes $x^2>-5+x$ when dividing both sides by $-1$

• Never multiply or divide both sides by a variable as this could be positive or negative

  • e.g. if $x>-2$ then $x^2>-2x$ is not always true

    • e.g. $x=-1$ satisfies $x>-2$ but not $x^2>-2x$

    • you get $1>2$

  • This means you can only multiply both sides by a terms that are always positive

    • Such as $x^2,\left|x\right|,{\mathrm{e}}^x$

• Taking reciprocals of positive values reverses the inequality

  • If $x,y>0$ then $x<y\Rightarrow \frac{1}{x}>\frac{1}{y}$

• Taking logarithms when the base is $0<a<1$ reverses the inequality

  • $0<x<y\Rightarrow {\log }_a\left(x\right)>{\log }_a\left(y\right)$

• The safest way to rearrange is simply to add and subtract terms to both sides

  • e.g. $-x^2<5-x$ becomes $0<5-x+x^2$