[tex]\it a)\ \ f(-2)\cdot f(4)=[(-2)^3+1]\sqrt{2\cdot4+1}=(-8+1)\sqrt{9}=-7\cdot3=-21\\ \\ \\ b)\\ \\ \left.\begin{aligned}f(0)=0^3+1=1\ \ \ \ \ \ \ \ \ \\ \\ \lim\limits_{x \nearrow0} f(x)= 0^3+1=1\ \ \ \\ \\ \lim\limits_{x \searrow 0} f(x)=\sqrt{2\cdot0+1}=1 \end{aligned}\right\} \Rightarrow f\ continu\breve a\ \hat{\imath}n\ x=0[/tex]
[tex]\it c)\\ \\ (p+1)(q+1)<0 \Rightarrow \begin{cases} \it I)\ p+1<0,\ \ q+1>0 \Rightarrow p<0,\ q>0\ \ \ (1)\\ \\ sau\\ \\ \it II)\ p+1>0,\ \ q+1<0 \Rightarrow p>0.\ \ q<0\ \ \ (2)\end{cases}\\ \\ \\ I)\ \ f(p) =p^3+1\\ \\ p +1<0 \Rightarrow p<-1 \Rightarrow p^3<-1 \Rightarrow p^3+1<0 \Rightarrow f(p)<0\ \ \ \ \ \ (3)\\ \\ f(q)=\sqrt{2q+1}\ >0\ \ \ \ \ \ (4)\\ \\ \\ (3),\ (4) \Rightarrow f(p)\cdot f(q)<0[/tex]
Analog se abordează cazul II)
