This note covers solving inequalities of the form \(g(x)\ge f(x)\), both graphically and analytically, including polynomial and rational cases. Answers are sets of \(x\)-values rather than single numbers, so careful sign analysis is essential.
Key ideas
To solve \(g(x)\ge f(x)\) means to find every \(x\) for which the graph of \(y=g(x)\) lies above or touches the graph of \(y=f(x)\). The solution is usually one or more intervals. The \(x\)-values where \(g(x)=f(x)\) are the boundaries between where the inequality is true and where it is false.
Graphical method. Plot \(y=f(x)\) and \(y=g(x)\) on your GDC. Find the intersection points. Then read off the intervals of \(x\) where the graph of \(g\) is above (or equal to) the graph of \(f\). Write the solution in inequality notation, giving the boundaries to 3 s.f.
Analytical method for polynomials. Rearrange to get zero on one side: \(g(x)-f(x)\ge 0\). Factorise the expression and find its roots. Then draw a sign diagram, or sketch the graph of the factorised expression, to see where it is positive or negative. A simple root changes the sign. A repeated root (squared factor) does not change the sign. With \(\ge\) or \(\le\), the roots themselves are included in the solution.
Analytical method for rational expressions. Never multiply both sides by an expression containing \(x\), because you do not know whether it is positive or negative. If it is negative, the inequality sign must reverse. Instead, move everything to one side and combine into a single fraction. For example, \(\dfrac{A}{B}\ge0\). Find the zeros of the numerator and of the denominator. The sign can only change at these values. Test each interval. The zeros of the numerator are included when the inequality is non-strict, but the zeros of the denominator are never included, because the expression is undefined there.
Writing the answer. Use inequality notation, such as \(x\le-2\) or \(x\gt 3\). The word "or" joins separate intervals. An answer like \(-2\le x\le 0\) is a single interval. Do not write "\(x\lt 1\) and \(x\gt 5\)", because no number satisfies both. Strict inequalities (\(\lt\), \(\gt\)) exclude the boundary, and non-strict ones (\(\le\), \(\ge\)) include it.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(g(x)\ge f(x)\iff g(x)-f(x)\ge0\) | Rearrange so that one side is zero before analysing signs. |
| Sign diagram | A number line marked with the zeros of each factor, showing where the expression is \(+\) or \(-\). |
| Simple root / repeated root | The sign changes at a simple root. It does not change at a repeated root. |
| \(\dfrac AB\ge0\) | Combine into one fraction. Exclude values where \(B=0\). |
| Multiplying by a negative | Reverses the inequality sign. Avoid multiplying by an unknown-sign expression. |
| Intersection points of \(f\) and \(g\) | Boundaries of the solution intervals when using the GDC. |
Worked examples
Example 1 Solve \(x^3-x^2\ge 6x\).
Move everything to one side: \(x^3-x^2-6x\ge0\).
Take out the common factor \(x\): \(x(x^2-x-6)\ge0\), and factorise the quadratic: \(x(x-3)(x+2)\ge0\).
The roots are \(x=-2\), \(x=0\) and \(x=3\), all simple, so the sign changes at each.
For large positive \(x\) all three factors are positive, so the product is positive when \(x\gt3\). Moving left, the sign changes at each root: negative for \(0\lt x\lt3\), positive for \(-2\lt x\lt0\), and negative for \(x\lt-2\).
Check with \(x=-1\): \((-1)(-4)(1)=4\gt0\). Check with \(x=1\): \((1)(-2)(3)=-6\lt 0\).
We want the expression to be positive or zero, and the roots are included.
Answer: \(-2\le x\le0\) or \(x\ge3\)
Example 2 Solve \(\dfrac{3}{x-1}\le x+1\).
Do not multiply by \(x-1\), because it may be negative. Move everything to one side: \(\dfrac{3}{x-1}-(x+1)\le0\).
Combine over a common denominator: \(\dfrac{3-(x+1)(x-1)}{x-1}\le0\), which is \(\dfrac{3-(x^2-1)}{x-1}=\dfrac{4-x^2}{x-1}\le0\).
Factorise the numerator: \(\dfrac{(2-x)(2+x)}{x-1}\le0\).
The critical values are \(x=-2\), \(x=1\) and \(x=2\). Test an \(x\)-value in each interval:
\(x=-3\): numerator \(4-9=-5\), denominator \(-4\), so the fraction is positive.
\(x=0\): numerator \(4\), denominator \(-1\), so the fraction is negative.
\(x=1.5\): numerator \(4-2.25=1.75\), denominator \(0.5\), so the fraction is positive.
\(x=3\): numerator \(4-9=-5\), denominator \(2\), so the fraction is negative.
We need the fraction to be negative or zero. It is zero at \(x=\pm2\), so those are included. At \(x=1\) the fraction is undefined, so that is excluded.
Answer: \(-2\le x\lt1\) or \(x\ge2\)
Example 3 Solve \(e^x\ge6-x^2\), giving the boundaries to 3 s.f.
This mixes an exponential and a quadratic, so use your GDC.
Plot \(y=e^x\) and \(y=6-x^2\). Use the intersection feature to find the two crossing points. The \(x\)-coordinates are \(-2.4315\ldots\) and \(1.3978\ldots\)
Now decide where the exponential graph is above the parabola. At \(x=0\), \(e^0=1\) and \(6-0=6\), so the parabola is higher, which means the inequality is false between the crossings.
For large positive \(x\), \(e^x\) grows far faster than \(6-x^2\) falls, so the exponential is above. For large negative \(x\), \(e^x\) is near 0 but \(6-x^2\) is very negative, so the exponential is also above.
Answer: \(x\le-2.43\) or \(x\ge1.40\)
Common errors
- Multiplying both sides of an inequality by an expression such as \(x-1\) without knowing its sign. This can reverse the inequality for some \(x\) and not others, giving wrong intervals.
- Giving only the roots as the answer instead of the intervals between or beyond them.
- Including the zeros of the denominator in the solution. The expression is undefined there.
- Writing a solution such as "\(x\lt-2\) and \(x\gt3\)". Use "or" for separate intervals.
- Assuming the sign alternates at every root. It does not change at a repeated root, such as \((x-2)^2\).
Exam tip: After finding your intervals, test one value from each in the original inequality. It takes a few seconds and catches reversed signs and wrongly included boundaries.
Check yourself
- Solve \(x^2+2x-3\lt0\).
Answer
Factorise: \((x+3)(x-1)\lt0\). The parabola opens upward, so the expression is negative between the roots. The solution is \(-3\lt x\lt1\). - Solve \(\dfrac{2x-1}{x+3}\ge1\).
Answer
Rearrange: \(\dfrac{2x-1}{x+3}-1=\dfrac{2x-1-(x+3)}{x+3}=\dfrac{x-4}{x+3}\ge0\). The critical values are \(-3\) and \(4\). The fraction is positive for \(x\lt-3\) and for \(x\gt4\), and zero at \(x=4\). The solution is \(x\lt-3\) or \(x\ge4\). The value \(x=-3\) is excluded. - Use your GDC to solve \(\ln x\le x-3\), giving boundaries to 3 s.f.
Answer
The graphs cross at \(x=0.0525\) and \(x=4.51\). The function \(\ln x\) is only defined for \(x\gt0\). At \(x=1\), \(\ln1=0\) is above \(1-3=-2\), so the inequality is false between the crossings. The solution is \(0\lt x\le0.0525\) or \(x\ge4.51\). - Solve \(x^3\ge x\).
Answer
\(x^3-x=x(x-1)(x+1)\ge0\). The roots are \(-1\), \(0\) and \(1\). The expression is positive for \(x\gt1\), negative for \(0\lt x\lt1\), positive for \(-1\lt x\lt0\), and negative for \(x\lt-1\). The solution is \(-1\le x\le0\) or \(x\ge1\).