Using Calculators to Solve Inequalities (Cambridge International IGCSE Maths)

Using Calculators to Solve Inequalities

How do I solve inequalities using my graphic display calculator?

• You can use a graphic display calculator to solve inequalities of the form

  • straight f_ 1 (x) greater than straight f_ 2 (x) or straight f_ 1 (x) less than straight f_ 2 (x)
  • $\mathrm{f} _{1} (x) \ge \mathrm{f} _{2} (x)$ or $\mathrm{f} _{1} (x) \le \mathrm{f} _{2} (x)$

• First, turn the inequality into an equation and solve $\text{straight} f_ 1 (x)=straight f_ 2 (x)$

  • Read the revision note on Using Calculators to Solve Equations to see how
  • The solutions to the equation $\text{straight} f_ 1 (x)=straight f_ 2 (x)$ are the x-coordinates of the points of intersection of the two graphs $y=straight f_ 1 (x)$ and $y=straight f_ 2 (x)$

• Then look at the regions (areas) enclosed between the two graphs and decide which curve is the top curve and which is the bottom curve

  • e.g. The solution to straight f_ 1 (x) greater than straight f_ 2 (x) are the ranges of x-values for which $y=straight f_ 1 (x)$ is the top curve and $y=straight f_ 2 (x)$ is the bottom curve

    • The top curve is greater than the bottom curve

  • Examples of different regions, R, are shown below

  • For example, to solve $x^{2}+3x+1>2x+1$ using a graphical method

    • First solve the "=" equation to get $x=-1$ and $x=0$
    • Then plot on your calculator and look for any regions where the quadratic is above (greater than) the line (quadratic on top, line on bottom)
    • This happens to the left of $x=-1$ and to the right of $x=0$
    • The solution is $x<-1$ or $x>0$

Exam Hint Remember to give your solutions as ranges of x values, using strict inequalities if the question uses strict inequalities!

Worked Example Use a graphical method to solve the inequality $x^{3}+2x^{3}+1\leq x+2$, giving your answers correct to 1 decimal place and showing a sketch of the graph.

Start a new graph on your calculator and enter straight f_ 1 (x)=x³+2x²+1 to generate the graph Add on the second graph of straight f_ 2 (x)=x+2

Select the option to "analyze" the graph (this may be labelled as G-Solv) Choose the option "intersections" to find the coordinates of the 3 points of intersection

The solutions to the equation are the x-coordinates of the points of intersection

The solutions to the equation $x^{3}+2x^{3}+1=x+2$ are

• $x=-2.2$,* $x=-0.6$ or $x=0.8$ to 1 d.p.

The inequality in the question is "cubic graph $\leq$ straight line graph" Look at regions where the cubic is the bottom curve and the line is above

left of x = -2.2 and between x = -0.6 and x = 0.8

The solutions are the ranges of x-values over which these regions lie Write your final answer using non-strict inequalities (to match the question)

Final answer: $x\leq-2.2$or$-0.6\leq x\leq0.8$to 1 d.p.