How Do You Solve \(g(x) \geq f(x)\) Graphically and Analytically?

Solve $g(x) \geq f(x)$ by finding where the graph of $g$ lies on or above the graph of $f$, or analytically by solving $g(x)-f(x) \geq 0$. Both methods identify the same set of $x$-values.

This is AHL 2.15: Solutions of inequalities in IB Mathematics: Analysis and Approaches HL.

Before using either method, state the domain and note any values where a function is undefined. Intersection points divide the domain into intervals. Test one value from each interval in (g(x)-f(x)): a positive result means $g(x)>f(x)$, zero gives equality, and a negative result means $g(x)<f(x)$. This prevents conclusions based only on the visual appearance of a graph.

For example, solve

$$x+2 \geq x^2.$$

Graphically, plot $g(x)=x+2$ and $f(x)=x^2$. Their intersections satisfy

$$x+2=x^2,$$

so they occur at $x=-1$ and $x=2$. Between these values, the line lies above the parabola. Therefore,

$$-1 \leq x \leq 2.$$

Analytically, rearrange consistently:

$$x^2-x-2 \leq 0,$$ $$(x-2)(x+1) \leq 0.$$

The product is non-positive between its roots, giving

$$x \in [-1,2].$$

For repeated roots or tangencies, the sign may not change across an intersection. Therefore, do not automatically alternate signs between consecutive roots. Check each interval directly. Also, use closed endpoints for $\geq$ or $\leq$, but open endpoints for $>$ or $<$, unless an endpoint is excluded from the original domain.

A common misconception is to report only the intersection points. The intersections are boundary values; the solution is usually one or more intervals. Because the inequality contains $\geq$, equality is allowed, so the endpoints are included.

In an IB exam, show the intersection equation or rearranged inequality before giving the interval. On Paper 1, expect an analytical method; on Paper 2, a GDC graph may be appropriate, but record the intersections and clearly state where $g(x) \geq f(x)$.