Question Number 20368 by Tinkutara last updated on 26/Aug/17
![If α is a real root of 2x^3 − 3x^2 + 6x + 6 = 0, then find [α] where [∙] denotes the greatest integer function.](https://www.tinkutara.com/question/Q20368.png)
$$\mathrm{If}\:\alpha\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{root}\:\mathrm{of}\:\mathrm{2}{x}^{\mathrm{3}} \:−\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:+\:\mathrm{6}\:=\:\mathrm{0}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\left[\alpha\right]\:\mathrm{where}\:\left[\centerdot\right]\:\mathrm{denotes}\:\mathrm{the} \\ $$$$\mathrm{greatest}\:\mathrm{integer}\:\mathrm{function}. \\ $$
Commented by ajfour last updated on 26/Aug/17
![f(x)=2x^3 −3x^2 +6x+6 f ′(x)=6x^2 −6x+6 =6(x−(1/2))^2 +(9/2) >0 Hence only one real root. f ′(x) > 0 and f(0)=6 so root α < 0 f(−1)=−5 , so α >−1 ⇒ −1< α < 0 ⇒ [α]=−1 .](https://www.tinkutara.com/question/Q20392.png)
$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{6} \\ $$$$\:{f}\:'\left({x}\right)=\mathrm{6}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{6}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{2}}\:>\mathrm{0} \\ $$$${Hence}\:{only}\:{one}\:{real}\:{root}. \\ $$$${f}\:'\left({x}\right)\:>\:\mathrm{0}\:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{6} \\ $$$${so}\:{root}\:\alpha\:<\:\mathrm{0} \\ $$$${f}\left(−\mathrm{1}\right)=−\mathrm{5}\:,\:{so}\:\alpha\:>−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:−\mathrm{1}<\:\alpha\:<\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\left[\alpha\right]=−\mathrm{1}\:. \\ $$
Commented by Tinkutara last updated on 26/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$