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Question Number 57012 by rahul 19 last updated on 28/Mar/19

Find the local minimum value of f(x) where f(x)= (x−(1/x))+((2/(x−(1/x)))) ?

$${Find}\:{the}\:{local}\:{minimum}\:{value}\:{of}\:{f}\left({x}\right) \\ $$$${where}\:{f}\left({x}\right)=\:\left({x}−\frac{\mathrm{1}}{{x}}\right)+\left(\frac{\mathrm{2}}{{x}−\frac{\mathrm{1}}{{x}}}\right)\:? \\ $$

Answered by mr W last updated on 29/Mar/19

let t=x−(1/x) if t>0: f(x)=t+(2/t)≥2(√(t×(2/t)))=2(√2)=local min if t<0: f(x)=t+(2/t)=−(−t+(2/(−t)))≤−2(√((−t)×(2/((−t)))))=−2(√2)=local max

$${let}\:{t}={x}−\frac{\mathrm{1}}{{x}} \\ $$$${if}\:{t}>\mathrm{0}: \\ $$$${f}\left({x}\right)={t}+\frac{\mathrm{2}}{{t}}\geqslant\mathrm{2}\sqrt{{t}×\frac{\mathrm{2}}{{t}}}=\mathrm{2}\sqrt{\mathrm{2}}={local}\:{min} \\ $$$${if}\:{t}<\mathrm{0}: \\ $$$${f}\left({x}\right)={t}+\frac{\mathrm{2}}{{t}}=−\left(−{t}+\frac{\mathrm{2}}{−{t}}\right)\leqslant−\mathrm{2}\sqrt{\left(−{t}\right)×\frac{\mathrm{2}}{\left(−{t}\right)}}=−\mathrm{2}\sqrt{\mathrm{2}}={local}\:{max} \\ $$

Commented by rahul 19 last updated on 30/Mar/19

thank you sir!

Commented by rahul 19 last updated on 30/Mar/19

Sir, how to do the same problem if f(x)= (x+(1/x))+((2/(x+(1/x)))) .

$${Sir},\:{how}\:{to}\:{do}\:{the}\:{same}\:{problem}\:{if}\: \\ $$$${f}\left({x}\right)=\:\left({x}+\frac{\mathrm{1}}{{x}}\right)+\left(\frac{\mathrm{2}}{{x}+\frac{\mathrm{1}}{{x}}}\right)\:. \\ $$

Commented by mr W last updated on 31/Mar/19

let t=x+(1/x) for x>0: t≥2 for x<0: t≤−2 f(t)=t+(2/t) f′(t)=1−(2/t^2 )>0⇒f(t) is strictly increasing if t≥2, i.e. x>0: sincs f(t) is strictly increasing ⇒min. =f(t=2)=3 if t≤−2, i.e. x<0: sincs f(t) is strictly increasing ⇒max. =f(t=−2)=−3

$${let}\:{t}={x}+\frac{\mathrm{1}}{{x}} \\ $$$${for}\:{x}>\mathrm{0}:\:{t}\geqslant\mathrm{2} \\ $$$${for}\:{x}<\mathrm{0}:\:{t}\leqslant−\mathrm{2} \\ $$$$ \\ $$$${f}\left({t}\right)={t}+\frac{\mathrm{2}}{{t}} \\ $$$${f}'\left({t}\right)=\mathrm{1}−\frac{\mathrm{2}}{{t}^{\mathrm{2}} }>\mathrm{0}\Rightarrow{f}\left({t}\right)\:{is}\:{strictly}\:{increasing} \\ $$$$ \\ $$$${if}\:{t}\geqslant\mathrm{2},\:{i}.{e}.\:{x}>\mathrm{0}: \\ $$$${sincs}\:{f}\left({t}\right)\:{is}\:{strictly}\:{increasing} \\ $$$$\Rightarrow{min}.\:={f}\left({t}=\mathrm{2}\right)=\mathrm{3}\: \\ $$$$ \\ $$$${if}\:{t}\leqslant−\mathrm{2},\:{i}.{e}.\:{x}<\mathrm{0}: \\ $$$${sincs}\:{f}\left({t}\right)\:{is}\:{strictly}\:{increasing} \\ $$$$\Rightarrow{max}.\:={f}\left({t}=−\mathrm{2}\right)=−\mathrm{3}\: \\ $$

Commented by rahul 19 last updated on 31/Mar/19

Sir ,in this case we cannot apply A.M≥G.M because t≥2 for x>0 unlike the above Q. where local minimum occured at t=(√2) ?

$${Sir}\:,{in}\:{this}\:{case}\:{we}\:{cannot}\:{apply} \\ $$$${A}.{M}\geqslant{G}.{M}\:{because}\:{t}\geqslant\mathrm{2}\:{for}\:{x}>\mathrm{0} \\ $$$${unlike}\:{the}\:{above}\:{Q}.\:{where}\:{local}\:{minimum} \\ $$$${occured}\:{at}\:{t}=\sqrt{\mathrm{2}}\:? \\ $$

Commented by mr W last updated on 31/Mar/19

that′s the reason, you are absolutely correct!

$${that}'{s}\:{the}\:{reason}, \\ $$$${you}\:{are}\:{absolutely}\:{correct}! \\ $$

Answered by MJS last updated on 28/Mar/19

f(x)=((x^4 +1)/(x(x^2 −1))) f′(x)=((x^6 −3x^4 −3x^2 +1)/(x^2 (x^2 −1)^2 )) x^6 −3x^4 −3x^2 +1=0 x=(√y) y^3 −3y^2 −3y+1=0 y=z+1 z^3 −6z−4=0 trying factors of −4 we find z_1 =−2 ⇒ y_1 =−1 ⇒ x_1 =±i (z+2)(z^2 −2z−2)=0 z_2 =1−(√3) ⇒ y_2 =2−(√3) ⇒ x_2 =±(((√6)/2)−((√2)/2)) z_3 =1+(√3) ⇒ y_3 =2+(√3) ⇒ x_3 =±(((√6)/2)+((√2)/2)) f′′(x)=((2(x^6 +9x^4 −3x^2 +1))/(x^3 (x^2 −1)^3 )) f′′(−((√6)/2)−((√2)/2))<0 f′′(−((√6)/2)+((√2)/2))>0 ⇒ min; f(x)=2(√2) f′′(((√6)/2)−((√2)/2))<0 f′′(((√6)/2)+((√2)/2))>0 ⇒ min ⇒ f(x)=2(√2)

$${f}\left({x}\right)=\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$${f}'\left({x}\right)=\frac{{x}^{\mathrm{6}} −\mathrm{3}{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${x}^{\mathrm{6}} −\mathrm{3}{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${x}=\sqrt{{y}} \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} −\mathrm{3}{y}+\mathrm{1}=\mathrm{0} \\ $$$${y}={z}+\mathrm{1} \\ $$$${z}^{\mathrm{3}} −\mathrm{6}{z}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{trying}\:\mathrm{factors}\:\mathrm{of}\:−\mathrm{4}\:\mathrm{we}\:\mathrm{find} \\ $$$${z}_{\mathrm{1}} =−\mathrm{2}\:\Rightarrow\:{y}_{\mathrm{1}} =−\mathrm{1}\:\Rightarrow\:{x}_{\mathrm{1}} =\pm\mathrm{i} \\ $$$$\left({z}+\mathrm{2}\right)\left({z}^{\mathrm{2}} −\mathrm{2}{z}−\mathrm{2}\right)=\mathrm{0} \\ $$$${z}_{\mathrm{2}} =\mathrm{1}−\sqrt{\mathrm{3}}\:\Rightarrow\:{y}_{\mathrm{2}} =\mathrm{2}−\sqrt{\mathrm{3}}\:\Rightarrow\:{x}_{\mathrm{2}} =\pm\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$${z}_{\mathrm{3}} =\mathrm{1}+\sqrt{\mathrm{3}}\:\Rightarrow\:{y}_{\mathrm{3}} =\mathrm{2}+\sqrt{\mathrm{3}}\:\Rightarrow\:{x}_{\mathrm{3}} =\pm\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$${f}''\left({x}\right)=\frac{\mathrm{2}\left({x}^{\mathrm{6}} +\mathrm{9}{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${f}''\left(−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)<\mathrm{0} \\ $$$${f}''\left(−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)>\mathrm{0}\:\Rightarrow\:\mathrm{min};\:{f}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${f}''\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)<\mathrm{0} \\ $$$${f}''\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)>\mathrm{0}\:\Rightarrow\:\mathrm{min}\:\Rightarrow\:{f}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{2}} \\ $$

Commented by rahul 19 last updated on 30/Mar/19

thank you sir!

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