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Question Number 12601 by @ANTARES_VY last updated on 26/Apr/17
y=(x^3 /3)+2x^2 −5x+7 find the critical points of the function
$$\boldsymbol{\mathrm{y}}=\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{7}\:\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{critical}}\:\:\boldsymbol{\mathrm{points}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{function}} \\ $$
Answered by mrW1 last updated on 26/Apr/17
y=(x^3 /3)+2x^2 −5x+7 =0 ⇒x=−8.16 y′=x^2 +4x−5=0 ⇒(x+5)(x−1)=0 ⇒x=−5 ⇒x=1 y′′=2x+4=0 ⇒x=−2
$$\boldsymbol{\mathrm{y}}=\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{7}\:=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{8}.\mathrm{16} \\ $$$$ \\ $$$${y}'={x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{5}\right)\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{5} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$$ \\ $$$${y}''=\mathrm{2}{x}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{2} \\ $$
Answered by ajfour last updated on 26/Apr/17
(dy/dx) = x^2 +4x−5 =(x+2)^2 −9 ⇒ (dy/dx)= 0 at x=−2±3 = −5,1 local maxima at x =−5 y(−5) =((121)/3) local minima at x =1 y(1) = ((13)/3) (d^2 y/dx^2 ) = 2x+4 ⇒ point of inflexion at x =−2 y(−2)=((67)/3) y=0 at x=α ; −9<α<−8 cannot evaluate.
$$\frac{{dy}}{{dx}}\:=\:{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{5}\:=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{9} \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}=\:\mathrm{0}\:{at}\:{x}=−\mathrm{2}\pm\mathrm{3}\:=\:−\mathrm{5},\mathrm{1} \\ $$$${local}\:{maxima}\:{at}\:{x}\:=−\mathrm{5}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}\left(−\mathrm{5}\right)\:=\frac{\mathrm{121}}{\mathrm{3}}\: \\ $$$${local}\:{minima}\:{at}\:{x}\:=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{13}}{\mathrm{3}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\mathrm{2}{x}+\mathrm{4}\:\:\: \\ $$$$\Rightarrow\:{point}\:{of}\:{inflexion}\:{at}\:{x}\:=−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}\left(−\mathrm{2}\right)=\frac{\mathrm{67}}{\mathrm{3}} \\ $$$${y}=\mathrm{0}\:\:{at}\:{x}=\alpha\:\:\:\:\:;\:\:\:\:−\mathrm{9}<\alpha<−\mathrm{8} \\ $$$${cannot}\:{evaluate}.\:\:\:\: \\ $$$$ \\ $$

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