Solving Inequalities from Graphs

Where the graphs cross

An inequality such as $x^{2} - 4 < x + 2$ compares two functions, $f\left(x\right) = x^{2} - 4$ 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 $x^{2} - 4 < x + 2$ for −2 < x < 3. At x = 0, for instance, −4 < 2.

Outside the crossings the parabola is above the line, so $x^{2} - 4 > x + 2$ 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): $x^{2} - 4 = x + 2$ gives $x^{2} - x - 6 = 0$, 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: $x^{2} - 4 \le x + 2$ for $-2 \le x \le 3$.

When the graphs do not cross

$x^{2} + 1 > x$ for every x. Moving x across gives $x^{2} - x + 1 > 0$, and $x^{2} - x + 1$ 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 $y = x^{2}$ at x = 1, since $x^{2} - 2x + 1 = \left(x - 1\right)^{2}$. So $x^{2} \le 2x - 1$ 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.