Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 174227 by mnjuly1970 last updated on 27/Jul/22

Q : Max_ _(x>1) ((( x^( 4) −x^( 2) )/(x^( 6) + 2x^( 3) − 1)) ) = ?

$$ \\ $$ $$\:\:\:{Q}\:: \\ $$ $$\:\:\:\:\:\:\:{Max}_{\underset{{x}>\mathrm{1}} {\:}} \:\left(\frac{\:{x}^{\:\mathrm{4}} −{x}^{\:\mathrm{2}} }{{x}^{\:\mathrm{6}} +\:\mathrm{2}{x}^{\:\mathrm{3}} −\:\mathrm{1}}\:\right)\:=\:? \\ $$ $$ \\ $$

Commented byinfinityaction last updated on 27/Jul/22

(1/6) ???

$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{6}}\:??? \\ $$

Commented bymnjuly1970 last updated on 27/Jul/22

yes Sir...

$${yes}\:\:{Sir}... \\ $$

Commented byinfinityaction last updated on 27/Jul/22

p = ((x−(1/x))/(x^3 −(1/x^3 )+2)) = ((x−(1/x))/((x−(1/x))(x^2 +(1/x^2 )+1)+2)) p = ((x−(1/x))/((x−(1/x)){(x−(1/x))^2 +3}+2)) let x−(1/x) = g p = (g/(g(g^2 +3)+2)) ⇒(1/(g^2 +(2/g)+3)) p = (1/(g^2 +(1/g)+(1/g)+3)) p_(max ) = (1/(3+3)) = (1/6)

$$\:\:{p}\:=\:\:\:\frac{{x}−\frac{\mathrm{1}}{{x}}}{{x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{2}}\:=\:\frac{{x}−\frac{\mathrm{1}}{{x}}}{\left({x}−\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right)+\mathrm{2}} \\ $$ $$\:\:\:\:\:{p}\:\:\:=\:\:\frac{{x}−\frac{\mathrm{1}}{{x}}}{\left({x}−\frac{\mathrm{1}}{{x}}\right)\left\{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{3}\right\}+\mathrm{2}} \\ $$ $$\:\:{let}\:{x}−\frac{\mathrm{1}}{{x}}\:=\:{g} \\ $$ $$\:\:\:\:{p}\:\:=\:\:\frac{{g}}{{g}\left({g}^{\mathrm{2}} +\mathrm{3}\right)+\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}}{{g}^{\mathrm{2}} +\frac{\mathrm{2}}{{g}}+\mathrm{3}} \\ $$ $$\:\:\:\:\:\:\:{p}\:\:=\:\:\frac{\mathrm{1}}{{g}^{\mathrm{2}} +\frac{\mathrm{1}}{{g}}+\frac{\mathrm{1}}{{g}}+\mathrm{3}} \\ $$ $$\:\:\:{p}_{{max}\:\:} =\:\:\frac{\mathrm{1}}{\mathrm{3}+\mathrm{3}}\:\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented bymnjuly1970 last updated on 27/Jul/22

very nice solution thanks alot sir

$$\:\:{very}\:{nice}\:{solution}\: \\ $$ $${thanks}\:{alot}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com