Solving an inequality

Quote:

Originally Posted by bobb12

Function $f: R \rightarrow R$ is a periodic function with period $T=6$ . For $x \in <-3,3>$ , the formula of this function is $f(x)=9-x^2$ . Solve the inequality $f(x) \le 5$ .

Note that \mathbb{R} gives $\mathbb{R}$

First solve the inequality for $x \in [-3,3]$ this gives

$9-x^2 \le 5 \iff 4-x^2 \le 0 \iff (2-x)(2+x) \le 0$ so the solution is

$[-3,-2] \cup [2,3]$

Since the function is periodic we know that $f(x)=f(x+6)$ So the inequality must must hold for when we shift right or left by 6 this gives

$[-3+6n,-2+6n] \cup [2 +6n,3 +6n]=[2 +6n,4 +6n], n \in \mathbb{Z}$