Solving Inequalities from Graphs

Where the graphs cross

An inequality such as $x^2-4<x+2$x² − 4 < x + 2 compares two functions, $f\left(x\right)=x^2-4$f(x) = x² − 4 and g(x) = x + 2. Draw both. Where the graphs cross, the two sides are equal.

These graphs cross at x = −2 and at x = 3. Everywhere else one graph is above the other, and which one is above can only change at a crossing.

Below and above

f(x) < g(x) means the graph of f is below the graph of g at that x. Between the crossings the parabola is below the line, so $x^2-4<x+2$x² − 4 < x + 2 for −2 < x < 3. At x = 0, for instance, −4 < 2.

Outside the crossings the parabola is above the line, so $x^2-4>x+2$x² − 4 > x + 2 for x < −2 or x > 3. At x = 4, 12 > 6. The answer has two parts, one on each side, and "or" joins them.

Finding the crossings exactly

A graph shows roughly where the crossings are. To find them exactly, solve f(x) = g(x): $x^2-4=x+2$x² − 4 = x + 2 gives $x^2-x-6=0$x² − x − 6 = 0, which factorizes as (x + 2)(x − 3) = 0, so x = −2 or x = 3.

With < or > the crossings are not in the answer, because there the two sides are equal. With $\le$≤ or $\ge$≥ they are: $x^2-4\le x+2$x² − 4 ≤ x + 2 for $-2\le x\le 3$−2 ≤ x ≤ 3.

When the graphs do not cross

$x^2+1>x$x² + 1 > x for every x. Moving x across gives $x^2-x+1>0$x² − x + 1 > 0, and $x^2-x+1$x² − x + 1 has discriminant 1 − 4 = −3, so it is never 0: the graphs never meet, and the parabola stays above the line.

A line can also just touch a parabola. y = 2x − 1 touches $y=x^2$y = x² at x = 1, since $x^2-2x+1={\left(x-1\right)}^2$x² − 2x + 1 = (x − 1)². So $x^2\le 2x-1$x² ≤ 2x − 1 holds only at x = 1.

The usual slips

Swapping the inside and the outside. Check one x: take a value between the crossings and see which side is bigger.

Losing one side of an outside answer. The parabola rises on both sides, so "x > 3" alone misses x < −2.

Dropping the crossings from $\le$≤ or $\ge$≥, where the two sides are equal and the inequality still holds.