Let f(x)=[x]^2-[x+3]-3,x R, where [.] is the greatest integer funtion.

Question

Let $f(x)=[x{]}^{2}-[x+3]-3,x\in R$, where [.] is the greatest integer funtion. Then

Accepted Answer

$\begin{gathered} \Rightarrow f(x)=0\Rightarrow [x]=3[x]=-2 \ \Rightarrow x\in [3,4)\cup [-2,-1)\Rightarrow f(x)=0 \ f(x)=[x{]}^{2}-[x+3]-3 \ =[{x}^{2}]-[x]-6=([x]-3)([x]+2) \end{gathered}$

$\begin{gathered} (2)\Rightarrow {\int }_{2}^{0}f(x)dx={\int }_{2}^{0}([x{]}^{2}-[x]-6)dx \ {\int }_{2}^{0}f(x)dx={\int }_{2}^{0}([x{]}^{2}-[x]-6)dx={\int }_{2}^{0}([x{]}^{2}-[x])dx-12=-12 \ (4)\Rightarrow f(x)0\Rightarrow [x]3\Rightarrow x<-2x\ge 4\Rightarrow x\in (-x,-2)0[4,\infty] \end{gathered}$

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