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Question Number 11473 by tawa last updated on 26/Mar/17

Show that the graph of  y = x^3  + 2x^2  + x + 1 , between x = − 1 and x = 2  lies entirely above the x − axis

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\:,\:\mathrm{between}\:\mathrm{x}\:=\:−\:\mathrm{1}\:\mathrm{and}\:\mathrm{x}\:=\:\mathrm{2} \\ $$$$\mathrm{lies}\:\mathrm{entirely}\:\mathrm{above}\:\mathrm{the}\:\mathrm{x}\:−\:\mathrm{axis} \\ $$

Commented by prakash jain last updated on 28/Mar/17

y=x^3 +2x^2 +x+1  =x(x^2 +2x+1)+1  =x(x+1)^2 +1  y>0 for x>0  −1≤x≤0  x(x+1)^2 >−1  x(x+1)^2 +1>0

$${y}={x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$$={x}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)+\mathrm{1} \\ $$$$={x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$${y}>\mathrm{0}\:{for}\:{x}>\mathrm{0} \\ $$$$−\mathrm{1}\leqslant{x}\leqslant\mathrm{0} \\ $$$${x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} >−\mathrm{1} \\ $$$${x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}>\mathrm{0} \\ $$$$ \\ $$

Answered by sm3l2996 last updated on 26/Mar/17

Commented by tawa last updated on 26/Mar/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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