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Question Number 7401 by txsnims last updated on 27/Aug/16

find the turning point on the curve y=((16)/x)+(x^3 /(3 )) and determine wether it is a point of maximum or minimum

$${find}\:{the}\:{turning}\:{point}\:{on}\:{the}\:{curve}\:{y}=\frac{\mathrm{16}}{{x}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}\:}\:\:{and}\:{determine}\:{wether}\:{it}\:{is}\:{a}\:{point}\:{of}\:{maximum}\:{or}\:{minimum} \\ $$$$ \\ $$

Answered by FilupSmith last updated on 27/Aug/16

y=16x^(−1) +3^(−1) x^3 turning point at y′=0 y^′ =−16x^(−2) +x^2 =0 x^2 =((16)/x^2 ) x^4 =16 (x^2 )^2 =(±4)^2 x=±2 for x∈R x=+2 y′′=32x^(−3) +2x y′′=32((1/8))+4(2) y′′>0 ∴minimum when x=2 x=−2 y′′=32((1/(−8)))+4(−2) y′′<0 ∴ maximum when x=−2

$${y}=\mathrm{16}{x}^{−\mathrm{1}} +\mathrm{3}^{−\mathrm{1}} {x}^{\mathrm{3}} \\ $$$$\mathrm{turning}\:\mathrm{point}\:\mathrm{at}\:{y}'=\mathrm{0} \\ $$$${y}^{'} =−\mathrm{16}{x}^{−\mathrm{2}} +{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{16}}{{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{4}} =\mathrm{16} \\ $$$$\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} =\left(\pm\mathrm{4}\right)^{\mathrm{2}} \\ $$$${x}=\pm\mathrm{2}\:\:\:\:\:\:\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$$$\: \\ $$$${x}=+\mathrm{2} \\ $$$${y}''=\mathrm{32}{x}^{−\mathrm{3}} +\mathrm{2}{x} \\ $$$${y}''=\mathrm{32}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)+\mathrm{4}\left(\mathrm{2}\right) \\ $$$${y}''>\mathrm{0} \\ $$$$\: \\ $$$$\therefore\mathrm{minimum}\:\mathrm{when}\:{x}=\mathrm{2} \\ $$$$ \\ $$$${x}=−\mathrm{2} \\ $$$${y}''=\mathrm{32}\left(\frac{\mathrm{1}}{−\mathrm{8}}\right)+\mathrm{4}\left(−\mathrm{2}\right) \\ $$$${y}''<\mathrm{0} \\ $$$$ \\ $$$$\therefore\:\mathrm{maximum}\:\mathrm{when}\:{x}=−\mathrm{2} \\ $$

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