Question Number 43748 by naka3546 last updated on 15/Sep/18
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Sep/18
$${f}\left({x}\right)=\mathrm{3}{x}^{\mathrm{3}} +{x}−\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=−{ve} \\ $$$${f}\left(\mathrm{1}\right)=+{ve} \\ $$$${f}\left(\mathrm{0}.\mathrm{5}\right)=−{ve} \\ $$$${f}\left(\mathrm{0}.\mathrm{6}\right)=+{ve} \\ $$$${so}\:\mathrm{0}.\mathrm{6}>{x}>\mathrm{0}.\mathrm{5} \\ $$$${f}\left(\mathrm{0}.\mathrm{55}\right)>+{ve} \\ $$$${f}\left(\mathrm{0}.\mathrm{53}\right)<−{ve} \\ $$$${x}\approx\mathrm{0}.\mathrm{54} \\ $$$$\mathrm{6}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2010} \\ $$$$\mathrm{6}\left(\mathrm{0}.\mathrm{54}\right)^{\mathrm{3}} −\left(\mathrm{0}.\mathrm{54}\right)^{\mathrm{2}} −\mathrm{3}\left(\mathrm{0}.\mathrm{54}\right)+\mathrm{2010} \\ $$$$\approx\mathrm{2009} \\ $$