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Question Number 121368 by abdullahquwatan last updated on 07/Nov/20

find minimun and maksimum y=((x^3 −1)/(x^2 −3))

$$\mathrm{find}\:\mathrm{minimun}\:\mathrm{and}\:\mathrm{maksimum} \\ $$$${y}=\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{3}} \\ $$

Answered by MJS_new last updated on 07/Nov/20

x^3 −1=0 ⇒ zero at x=1 x^2 −3≠0 ⇒ asymptotes x=±(√3) ⇒ −∞<y<+∞ ((x^3 −1)/(x^2 −3))=x+((3x−1)/(x^2 −3)) ⇒ asymptote y=x y′=((x(x^3 −9x+2))/((x^2 −3)^2 )) y′′=((6(x^3 −x^2 +9x−1))/((x^2 −3)^3 )) solving y′=0 and testing y′′ { ((<0 (maximum))),((=0 (flat point))),((>0 (minimum))) :} leads to x_1 ≈−3.10548 local max with y≈−4.65822 x_2 =0 local minimum with y=(1/3) x_3 ≈.223462 local maximum with y≈.335193 x_4 ≈2.88202 local minimum with y≈4.32303

$${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\:\mathrm{zero}\:\mathrm{at}\:{x}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}\neq\mathrm{0}\:\Rightarrow\:\mathrm{asymptotes}\:{x}=\pm\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:−\infty<{y}<+\infty \\ $$$$\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{3}}={x}+\frac{\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{3}}\:\Rightarrow\:\mathrm{asymptote}\:{y}={x} \\ $$$${y}'=\frac{{x}\left({x}^{\mathrm{3}} −\mathrm{9}{x}+\mathrm{2}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${y}''=\frac{\mathrm{6}\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{1}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{3}} } \\ $$$$\mathrm{solving}\:{y}'=\mathrm{0}\:\mathrm{and}\:\mathrm{testing}\:{y}''\:\begin{cases}{<\mathrm{0}\:\left(\mathrm{maximum}\right)}\\{=\mathrm{0}\:\left(\mathrm{flat}\:\mathrm{point}\right)}\\{>\mathrm{0}\:\left(\mathrm{minimum}\right)}\end{cases} \\ $$$$\mathrm{leads}\:\mathrm{to} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{3}.\mathrm{10548}\:\mathrm{local}\:\mathrm{max}\:\mathrm{with}\:{y}\approx−\mathrm{4}.\mathrm{65822} \\ $$$${x}_{\mathrm{2}} =\mathrm{0}\:\mathrm{local}\:\mathrm{minimum}\:\mathrm{with}\:{y}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}_{\mathrm{3}} \approx.\mathrm{223462}\:\mathrm{local}\:\mathrm{maximum}\:\mathrm{with}\:{y}\approx.\mathrm{335193} \\ $$$${x}_{\mathrm{4}} \approx\mathrm{2}.\mathrm{88202}\:\mathrm{local}\:\mathrm{minimum}\:\mathrm{with}\:{y}\approx\mathrm{4}.\mathrm{32303} \\ $$

Commented by abdullahquwatan last updated on 07/Nov/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by MJS_new last updated on 07/Nov/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

Answered by TANMAY PANACEA last updated on 07/Nov/20

(dy/dx)=(((x^2 −3)(3x^2 )−(x^3 −1)2x)/((x^2 −3)^2 )) (dy/dx)=((3x^4 −9x^2 −2x^4 +2x)/((x^2 −3)^2 ))=((x^4 −9x^2 +2x)/((x^2 −3)^2 )) for max/min (dy/dx)=0 x(x^3 −9x+2)=0 x=0 and 0.223, 2.882 when x>2.882 (dy/dx)>0 when2.882 >x>0.223 (dy/dx)<0 when0.223>x>0 (dy/dx)>0 when x<0 (dy/dx)<0 sign of (dy/dx) changes from −ve to +ve when curve cross x=0 so at x=0 given function minimum y_(x=0) =(1/3)(minimum)=(1/3) sign change from −ve to +ve when cross x=0.223 so maximum y_(x=0.223) =(((0.223)^3 −1)/((0.223)^2 −3))(maximum)= when 2.882>x (dy/dx)<0 when x>2.882 (dy/dx)>0 sign change from −ve to +ve so minimum y_(x=2.882) =(((2.882)^3 −1)/((2.882)^2 −3))( minimum)=

$$\frac{{dy}}{{dx}}=\frac{\left({x}^{\mathrm{2}} −\mathrm{3}\right)\left(\mathrm{3}{x}^{\mathrm{2}} \right)−\left({x}^{\mathrm{3}} −\mathrm{1}\right)\mathrm{2}{x}}{\left({x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{3}{x}^{\mathrm{4}} −\mathrm{9}{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{4}} +\mathrm{2}{x}}{\left({x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} }=\frac{{x}^{\mathrm{4}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{2}{x}}{\left({x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${for}\:{max}/{min}\:\frac{{dy}}{{dx}}=\mathrm{0}\:\:{x}\left({x}^{\mathrm{3}} −\mathrm{9}{x}+\mathrm{2}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:{and}\:\:\mathrm{0}.\mathrm{223},\:\mathrm{2}.\mathrm{882} \\ $$$${when}\:{x}>\mathrm{2}.\mathrm{882}\:\:\:\:\frac{{dy}}{{dx}}>\mathrm{0} \\ $$$${when}\mathrm{2}.\mathrm{882}\:>{x}>\mathrm{0}.\mathrm{223}\:\:\:\:\:\:\frac{{dy}}{{dx}}<\mathrm{0} \\ $$$${when}\mathrm{0}.\mathrm{223}>{x}>\mathrm{0}\:\:\:\:\:\:\frac{{dy}}{{dx}}>\mathrm{0} \\ $$$${when}\:{x}<\mathrm{0}\:\:\frac{{dy}}{{dx}}<\mathrm{0} \\ $$$${sign}\:{of}\:\frac{{dy}}{{dx}}\:{changes}\:{from}\:−{ve}\:{to}\:+{ve}\:\:{when}\:{curve}\:{cross}\:{x}=\mathrm{0} \\ $$$${so}\:{at}\:{x}=\mathrm{0}\:\:{given}\:{function}\:{minimum} \\ $$$${y}_{{x}=\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{3}}\left({minimum}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${sign}\:{change}\:{from}\:−{ve}\:{to}\:+{ve}\:{when}\:{cross}\:{x}=\mathrm{0}.\mathrm{223} \\ $$$${so}\:{maximum}\:{y}_{{x}=\mathrm{0}.\mathrm{223}} =\frac{\left(\mathrm{0}.\mathrm{223}\right)^{\mathrm{3}} −\mathrm{1}}{\left(\mathrm{0}.\mathrm{223}\right)^{\mathrm{2}} −\mathrm{3}}\left({maximum}\right)= \\ $$$${when}\:\mathrm{2}.\mathrm{882}>{x}\:\:\frac{{dy}}{{dx}}<\mathrm{0} \\ $$$${when}\:{x}>\mathrm{2}.\mathrm{882}\:\:\frac{{dy}}{{dx}}>\mathrm{0}\: \\ $$$${sign}\:{change}\:{from}\:−{ve}\:{to}\:+{ve}\: \\ $$$${so}\:{minimum} \\ $$$${y}_{{x}=\mathrm{2}.\mathrm{882}} \:=\frac{\left(\mathrm{2}.\mathrm{882}\right)^{\mathrm{3}} −\mathrm{1}}{\left(\mathrm{2}.\mathrm{882}\right)^{\mathrm{2}} −\mathrm{3}}\left(\:{minimum}\right)= \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by TANMAY PANACEA last updated on 07/Nov/20

pls calculate the[value

$$\boldsymbol{{pls}}\:\boldsymbol{{calculate}}\:\boldsymbol{{the}}\left[\boldsymbol{{value}}\right. \\ $$

Commented by abdullahquwatan last updated on 07/Nov/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by TANMAY PANACEA last updated on 07/Nov/20

most welcome

$${most}\:{welcome} \\ $$

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