Question Number 11473 by tawa last updated on 26/Mar/17
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\:,\:\mathrm{between}\:\mathrm{x}\:=\:−\:\mathrm{1}\:\mathrm{and}\:\mathrm{x}\:=\:\mathrm{2} \\ $$$$\mathrm{lies}\:\mathrm{entirely}\:\mathrm{above}\:\mathrm{the}\:\mathrm{x}\:−\:\mathrm{axis} \\ $$
Commented by prakash jain last updated on 28/Mar/17
$${y}={x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$$={x}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)+\mathrm{1} \\ $$$$={x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$${y}>\mathrm{0}\:{for}\:{x}>\mathrm{0} \\ $$$$−\mathrm{1}\leqslant{x}\leqslant\mathrm{0} \\ $$$${x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} >−\mathrm{1} \\ $$$${x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}>\mathrm{0} \\ $$$$ \\ $$
Answered by sm3l2996 last updated on 26/Mar/17
Commented by tawa last updated on 26/Mar/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$