# Solving an inequality

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• Mar 29th 2011, 06:50 AM
bobb12
Solving an inequality
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 $f(x) \le 5$.
• Mar 29th 2011, 07:06 AM
TheEmptySet
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

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