# Sketching non-linear inequalities

How do I sketch non-linear inequalities?

For instance, how do I sketch $x^3<1,x\in\mathbb{R}$?

For linear inequalities, I just sketch the equalities and wonder about which side of line the inequality is talking about.

How do I sketch nonlinear inequalities? I can ofcourse sketch the equality but I don't know where to go from there.

• Do you know something about the real function $f(x)=x^3$? – Sigur Nov 9 '12 at 0:51
• "I just sketch the equalities and wonder about which side of line the inequality is talking about." This sentence makes it sound as though you are referring to shading the entire plane above or below a particular line. This is appropriate for two-variable inequalities (such as $2x + 1 < y$). The non-linear example you give involves only a single variable, so you would not shade a portion of the plane. Rather, you would look only for certain $x$-values that satisfy the inequality (see my answer). – Austin Mohr Nov 9 '12 at 1:10

## 1 Answer

One way is to rewrite the inequality as $x^3 - 1 < 0$. The values of $x$ that satisfy the inequalities (both this modified inequality and the original one) are precisely those for which the graph of $x^3 - 1$ lies strictly below the $x$-axis. The solution set for your example is therefore all $x$ belonging to the interval $(-\infty, 1)$. You could "sketch" the solution set as I have done in red. • So, if I understand correctly, $f(x) < c$ is the intersection of the epigraph of f and the area defined by $x<c$? – Inquest Nov 9 '12 at 1:42
• @Inquest That is not true. Such an intersection would be two-dimensional. As you have only one variable $x$, the solution set can be at most one-dimensional. That is, the solution set consists of real numbers, not ordered pairs of real numbers. Think of the inequality as a test that each real number can either pass or fail. The number $-1$ belongs to the solution set, because $(-1)^3 - 1 < 0$ is a true statement. The number 2 does not belong to the solution set because $2^3 - 1 < 0$ is a false statement. – Austin Mohr Nov 9 '12 at 1:46
• @Inquest Let me stress again that it is the real number $-1$ that belongs to the solution set, not the ordered pair $(-1,-2)$, nor any other kind of ordered pair. An ordered pair would belong to the solution set of the inequality $x^3 - 1 < y$, for example. – Austin Mohr Nov 9 '12 at 2:00