Function $\displaystyle f: R \rightarrow R$ is a periodic function with period $\displaystyle T=6$. For $\displaystyle x \in <-3,3>$, the formula of this function is $\displaystyle f(x)=9-x^2$. Solve the inequality $\displaystyle f(x) \le 5$.
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Function $\displaystyle f: R \rightarrow R$ is a periodic function with period $\displaystyle T=6$. For $\displaystyle x \in <-3,3>$, the formula of this function is $\displaystyle f(x)=9-x^2$. Solve the inequality $\displaystyle f(x) \le 5$.
Note that \mathbb{R} gives $\displaystyle \mathbb{R}$Quote:
Originally Posted by bobb12
First solve the inequality for $\displaystyle x \in [-3,3]$ this gives
$\displaystyle 9-x^2 \le 5 \iff 4-x^2 \le 0 \iff (2-x)(2+x) \le 0 $ so the solution is
$\displaystyle [-3,-2] \cup [2,3]$
Since the function is periodic we know that $\displaystyle f(x)=f(x+6)$ So the inequality must must hold for when we shift right or left by 6 this gives
$\displaystyle [-3+6n,-2+6n] \cup [2 +6n,3 +6n]=[2 +6n,4 +6n], n \in \mathbb{Z}$
