2.15 Solving inequalitiesIB Maths: Analysis and Approaches HL: Revision notes

Solving $g(x)\ge f(x)$ means finding every $x$ for which the graph of $y=g(x)$ is on or above the graph of $y=f(x)$. The key steps are the same every time:

1. Find the critical values: where $g(x)=f(x)$, and where either function is undefined.
2. Decide the sign of $g(x)-f(x)$ on each interval between critical values.
3. Write the solution set, taking care with strict ($>$) versus non-strict ($\ge$) endpoints.

It is often easiest to rewrite the problem as $h(x)=g(x)-f(x)\ge0$ and study the sign of $h$.

For a polynomial written in factors, the sign only changes at simple roots. With a positive leading coefficient:

• a quadratic with roots $a<b$ is negative between the roots and positive outside;
• a cubic with roots $a<b<c$ is negative for $x<a$, positive on $(a,b)$, negative on $(b,c)$, positive for $x>c$.

Example: $(x+2)(x-1)(x-4)>0$ gives $-2<x<1$ or $x>4$.

A repeated root does not change the sign: $(x-2)^{2}(x-8)\ge0$ gives $x=2$ or $x\ge8$.

If the cubic is not factorised, use the factor theorem to find one root, then divide to get a quadratic.

If you multiply both sides of an inequality by an expression such as $(x-2)$, the inequality reverses when that expression is negative. Two safe methods:

• Bring everything to one side as a single fraction and study the signs of numerator and denominator.
• Multiply both sides by the square $(x-2)^{2}$, which is positive for $x\neq2$.

Example: $\dfrac{x}{x-2}\le x-4\iff\dfrac{x^{2}-7x+8}{x-2}\ge0$. Critical values: $\frac{7\pm\sqrt{17}}{2}$ and $2$. The solution is $\frac{7-\sqrt{17}}{2}\le x<2$ or $x\ge\frac{7+\sqrt{17}}{2}$. The value $x=2$ is never included, because $f(2)$ is undefined.

For inequalities with no algebraic method, such as $\dfrac{x}{x-2}>\mathrm{e}^{x-3}$, use the GDC:

• graph both sides and find their intersections, or
• graph $h(x)=g(x)-f(x)$ and find its zeros.

Then read off where one graph is above the other, and state the answer with boundaries to 3 significant figures, e.g. $2<x<3.76$. Always include any restriction on the domain (here $x>2$) and any vertical asymptote as a boundary.

In modelling problems, the model's domain restricts the answer. For a box of volume $V(x)=x(12-2x)^{2}$, $0<x<6$: $V(x)\ge100\iff(x-1)(x^{2}-11x+25)\ge0.$ The cubic is non-negative for $1\le x\le\frac{11-\sqrt{21}}{2}$ or $x\ge\frac{11+\sqrt{21}}{2}$, but the second interval is outside the domain, so the answer is $1\le x\le\frac{11-\sqrt{21}}{2}$ (about $3.21$).