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

$$\mathrm{Find}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\frac{(5+\mathrm{x}){(2+\mathrm{x})}^2}{\mathrm{x}(1+\mathrm{x})}(\mathrm{x}\ne -1,0)$$

Commented by MJS_new last updated on 03/Oct/20

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

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

$$\mathrm{Yes}$$

Answered by MJS_new last updated on 03/Oct/20

$$y=\frac{{(x+2)}^2(x+5)}{x(x+1)}$$$$y^{\prime }=\frac{(x+2)(x^3-15x-10)}{x^2{(x+1)}^2}$$$$y^{″}=\frac{8(4x^3+15x^2+15x+5)}{x^3{(x+1)}^3}$$$$y^{\prime }=0$$$$\Rightarrow$$$$x_1\approx -3.48261\wedge y\approx .385778;y^{″}<0\Rightarrow \mathrm{local}\max$$$$x_2=-2\wedge y=0;y^{″}>0\Rightarrow \mathrm{local}\min$$$$x_3\approx -.688417\wedge y\approx -34.5782;y^{″}<0\Rightarrow \mathrm{local}\max$$$$x_4\approx 4.17103\wedge y\approx 16.1925;y^{″}>0\Rightarrow \mathrm{local}\min$$

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

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

Commented by MJS_new last updated on 03/Oct/20

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

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