Use ordered roots and interval signs to solve a cubic inequality. Reflect only the negative part of a quadratic graph in the x-axis.
Notes and key methods
For three distinct simple real roots, a cubic changes sign at each root. The leading coefficient fixes the sign to the far right. A test value in each interval can confirm the sign chart.
A non-strict inequality includes its zero boundaries. For \(\displaystyle y=|P(x)|\), reflect only the intervals where the cubic is negative; these roots remain the meeting points of the reflected pieces.
To obtain \(\displaystyle y=|f(x)|\), keep portions with \(\displaystyle f(x)\ge 0\) and reflect portions with f(x)<0 upwards. The x-coordinates of zeros stay fixed. The resulting graph is nonnegative.
For \(f(x)=(x-h)^2-d^2\), the portion between h−d and h+d is reflected. It has a local maximum at \(\displaystyle x=h\); the two zeros become sharp local minima. Adding k moves the entire result vertically.
Check the conditions
The three roots in these problems are distinct and simple.
Include or exclude roots according to the inequality symbol.
Determine the sign of the original quadratic first.
Reflection changes y-values, not x-values.
Three worked examples
Example 1 · Cubic sign charts
\(\displaystyle P(x)=\)\(\left(x - 10\right) \left(x - 6\right) \left(x - 3\right)\). Use a sign chart to analyse P.
Use the labelled scales alongside the numerical method.
Solve \(\displaystyle P(x)\ge 0\).
Solve P(x)<0.
Find the largest root.
Evaluate P at the midpoint of the last two roots.
Show the reasoning
(a)
The signs from left to right are −,+,−,+; include all three zero boundaries.
