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Question Number 116348 by Dwaipayan Shikari last updated on 03/Oct/20

Find the minimum value of (((5+x)(2+x)^2 )/(x(1+x))) (x≠−1,0)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\frac{\left(\mathrm{5}+\mathrm{x}\right)\left(\mathrm{2}+\mathrm{x}\right)^{\mathrm{2}} }{\mathrm{x}\left(\mathrm{1}+\mathrm{x}\right)}\:\:\left(\mathrm{x}\neq−\mathrm{1},\mathrm{0}\right) \\ $$

Commented by MJS_new last updated on 03/Oct/20

−∞<(((5+x)(2+x)^2 )/(x(1+x)))<∞ ⇒ minimum is −∞ or do you want local minima?

$$−\infty<\frac{\left(\mathrm{5}+{x}\right)\left(\mathrm{2}+{x}\right)^{\mathrm{2}} }{{x}\left(\mathrm{1}+{x}\right)}<\infty \\ $$$$\Rightarrow\:\mathrm{minimum}\:\mathrm{is}\:−\infty \\ $$$$\mathrm{or}\:\mathrm{do}\:\mathrm{you}\:\mathrm{want}\:\mathrm{local}\:\mathrm{minima}? \\ $$

Commented by Dwaipayan Shikari last updated on 03/Oct/20

Yes

$$\mathrm{Yes} \\ $$

Answered by MJS_new last updated on 03/Oct/20

y=(((x+2)^2 (x+5))/(x(x+1))) y′=(((x+2)(x^3 −15x−10))/(x^2 (x+1)^2 )) y′′=((8(4x^3 +15x^2 +15x+5))/(x^3 (x+1)^3 )) y′=0 ⇒ x_1 ≈−3.48261∧y≈.385778; y′′<0 ⇒ local max x_2 =−2∧y=0; y′′>0 ⇒ local min x_3 ≈−.688417∧y≈−34.5782; y′′<0 ⇒ local max x_4 ≈4.17103∧y≈16.1925; y′′>0 ⇒ local min

$${y}=\frac{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \left({x}+\mathrm{5}\right)}{{x}\left({x}+\mathrm{1}\right)} \\ $$$${y}'=\frac{\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{3}} −\mathrm{15}{x}−\mathrm{10}\right)}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${y}''=\frac{\mathrm{8}\left(\mathrm{4}{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{5}\right)}{{x}^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${y}'=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{3}.\mathrm{48261}\wedge{y}\approx.\mathrm{385778};\:{y}''<\mathrm{0}\:\Rightarrow\:\mathrm{local}\:\mathrm{max} \\ $$$${x}_{\mathrm{2}} =−\mathrm{2}\wedge{y}=\mathrm{0};\:{y}''>\mathrm{0}\:\Rightarrow\:\mathrm{local}\:\mathrm{min} \\ $$$${x}_{\mathrm{3}} \approx−.\mathrm{688417}\wedge{y}\approx−\mathrm{34}.\mathrm{5782};\:{y}''<\mathrm{0}\:\Rightarrow\:\mathrm{local}\:\mathrm{max} \\ $$$${x}_{\mathrm{4}} \approx\mathrm{4}.\mathrm{17103}\wedge{y}\approx\mathrm{16}.\mathrm{1925};\:{y}''>\mathrm{0}\:\Rightarrow\:\mathrm{local}\:\mathrm{min} \\ $$

Commented by Dwaipayan Shikari last updated on 03/Oct/20

Thanking you

$$\mathrm{Thanking}\:\mathrm{you} \\ $$

Commented by MJS_new last updated on 03/Oct/20

you′re welcome the exact values of the roots if the 3^(rd) degree polynomial are not useful in this case

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$$$\mathrm{the}\:\mathrm{exact}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{if}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree} \\ $$$$\mathrm{polynomial}\:\mathrm{are}\:\mathrm{not}\:\mathrm{useful}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$

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