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Question Number 20368 by Tinkutara last updated on 26/Aug/17

$$\mathrm{If}\alpha \mathrm{is}\mathrm{a}\mathrm{real}\mathrm{root}\mathrm{of}2x^3-3x^2+6x+6=0,$$$$\mathrm{then}\mathrm{find}\left[\alpha \right]\mathrm{where}[\cdot ]\mathrm{denotes}\mathrm{the}$$$$\mathrm{greatest}\mathrm{integer}\mathrm{function}.$$

Commented by ajfour last updated on 26/Aug/17

$$f\left(x\right)=2x^3-3x^2+6x+6$$$$f{}^{\prime }\left(x\right)=6x^2-6x+6$$$$=6{(x-\frac{1}{2})}^2+\frac{9}{2}>0$$$$Henceonlyonerealroot.$$$$f{}^{\prime }\left(x\right)>0andf\left(0\right)=6$$$$soroot\alpha <0$$$$f(-1)=-5,so\alpha >-1$$$$\Rightarrow -1<\alpha <0$$$$\Rightarrow \left[\alpha \right]=-1.$$

Commented by Tinkutara last updated on 26/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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