Using Calculators to Solve Inequalities (Cambridge (CIE) IGCSE International Maths)

Revision Note

Author

Mark Curtis

Expertise

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
- $f_{1} (x) > f_{2} (x)$ or $f_{1} (x) < f_{2} (x)$
- $f_{1} (x) \geq f_{2} (x)$ or $f_{1} (x) \leq f_{2} (x)$
First, turn the inequality into an equation and solve $f_{1} (x) = f_{2} (x)$
- Read the revision note on Using Calculators to Solve Equations to see how
- The solutions to the equation $f_{1} (x) = f_{2} (x)$ are the x-coordinates of the points of intersection of the two graphs $y = f_{1} (x)$ and $y = f_{2} (x)$
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 $f_{1} (x) > f_{2} (x)$ are the ranges of x-values for which $y = f_{1} (x)$ is the top curve and $y = f_{2} (x)$ 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 $x^{2} + 3 x + 1 > 2 x + 1$ 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 $x > 0$

Points of intersection between a curve and a line

Exam Tip

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 $x^{3} + 2 x^{3} + 1 \leq x + 2$ , giving your answers correct to 1 decimal place and showing a sketch of the graph.

Start a new graph on your calculator and enter $f_{1} (x) = x^{3} + 2 x^{2} + 1$ to generate the graph
Add on the second graph of $f_{2} (x) = x + 2$

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

Drawing a cubic graph and a straight line graph on a graphic display calculator to find the points of intersection

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

$x \leq - 2.2$ or $- 0.6 \leq x \leq 0.8$ to 1 d.p.

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