Solving Inequalities

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Now that we understand how a sign table is constructed, we can use these table to solve inequalities.

Let’s try some examples:

Solve the inequality $(x-2)(x+3) < 0$

We will solve this inequality by constructing a sign table for $f(x)=(x-2)(x+3)$ and using this information to find the solution set.

From the previous section, we know that the sign table for $f(x)$ is given by

We are looking for all values of $x$ such that $f(x)<0$ . Checking the sign table, we see that the solution set (in interval notation) is

$[-3,2]$ $(-\infty ,-3]\cup [2,\infty )$ $(-\infty ,-3)\cup (2,\infty )$ $[-3,2)$ $(-3,2)$

Solve the inequality $\frac {(x-4)(x+4)}{(x+2)} \geq 0$

We will solve this inequality by constructing a sign table for $f(x) = \frac {(x-4)(x+4)}{(x+2)}$ and using this information to find the solution set. From the previous section, we know that the sign table for $f(x)$ is given by

We are looking for all values of $x$ such that $f(x)\geq 0$ . If we first limit ourselves to the points where $f(x)>0$ then we see that

$f(x) > 0$ if $x$ is in the interval $(-4,-2) \text { or } (4,\infty )$ Finally, as we need $f(x) \geq 0$ , we also have to include the points where $f(x)$ =0, namely $x=-4$ and $x=4$ . Therefore our solution set (in interval notation) is

$(-\infty ,-4]\cup [-2,4)$ $(-4,-2)\cup (4,\infty )$ $(-4,-2]\cup (4,\infty )$ $(-\infty ,-4)\cup (-2,4)$ $(-4,-2]\cup [4,\infty )$

We’ll meet sign tables again when we discuss how calculus helps us to find analytic information about a function.