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 100370 by Rio Michael last updated on 26/Jun/20

Find the maximum value of  f(x) = (3/(2cosh (ln x) + 3))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\:{f}\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2cosh}\:\left(\mathrm{ln}\:{x}\right)\:+\:\mathrm{3}} \\ $$

Commented by bemath last updated on 26/Jun/20

y_(max) = 0.6

$$\mathrm{y}_{\mathrm{max}} =\:\mathrm{0}.\mathrm{6} \\ $$

Commented by mr W last updated on 26/Jun/20

maximum f(x) means mininum cosh (ln x),  which is 1 when ln x=0 or x=1.  max. f(x)=f(1)=(3/(2×1+3))=(3/5)

$${maximum}\:{f}\left({x}\right)\:{means}\:{mininum}\:\mathrm{cosh}\:\left(\mathrm{ln}\:{x}\right), \\ $$$${which}\:{is}\:\mathrm{1}\:{when}\:\mathrm{ln}\:{x}=\mathrm{0}\:{or}\:{x}=\mathrm{1}. \\ $$$${max}.\:{f}\left({x}\right)={f}\left(\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{2}×\mathrm{1}+\mathrm{3}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$

Answered by MJS last updated on 26/Jun/20

cosh ln x =(1/2)(x+(1/x))  f(x)=((3x)/(x^2 +3x+1))  f′(x)=−((3(x^2 −1))/((x^2 +3x+1)^2 ))  f′(x)=0 ⇒ x=±1  f′′(x)=((6(x^3 −3x−3))/((x^2 +3x+1)^3 ))  f′′(−1)=6>0 ⇒ min at x=−1; y=3  f′′(1)=−(6/(25))<0 ⇒ max at x=1; y=(3/5)  but these are only local  −∞<f(x)<+∞  f(x) is not defined for x=−(3/2)±((√5)/2)

$$\mathrm{cosh}\:\mathrm{ln}\:{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{3}{x}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}} \\ $$$${f}'\left({x}\right)=−\frac{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${f}'\left({x}\right)=\mathrm{0}\:\Rightarrow\:{x}=\pm\mathrm{1} \\ $$$${f}''\left({x}\right)=\frac{\mathrm{6}\left({x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{3}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${f}''\left(−\mathrm{1}\right)=\mathrm{6}>\mathrm{0}\:\Rightarrow\:\mathrm{min}\:\mathrm{at}\:{x}=−\mathrm{1};\:{y}=\mathrm{3} \\ $$$${f}''\left(\mathrm{1}\right)=−\frac{\mathrm{6}}{\mathrm{25}}<\mathrm{0}\:\Rightarrow\:\mathrm{max}\:\mathrm{at}\:{x}=\mathrm{1};\:{y}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{but}\:\mathrm{these}\:\mathrm{are}\:\mathrm{only}\:\mathrm{local} \\ $$$$−\infty<{f}\left({x}\right)<+\infty \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:{x}=−\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Commented by bemath last updated on 26/Jun/20

sir cosh (ln x) = (1/2)(x+(1/x)) it is from series?

$$\mathrm{sir}\:\mathrm{cosh}\:\left(\mathrm{ln}\:{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{{x}}\right)\:{it}\:{is}\:{from}\:{series}? \\ $$

Commented by MJS last updated on 26/Jun/20

no; from definition  cosh x =((e^x +e^(−x) )/2)  sinh x =((e^x −e^(−x) )/2)  cos x =((e^(ix) +e^(ix) )/2)  sin x =((e^(ix) −e^(−ix) )/(2i))

$$\mathrm{no};\:\mathrm{from}\:\mathrm{definition} \\ $$$$\mathrm{cosh}\:{x}\:=\frac{\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} }{\mathrm{2}} \\ $$$$\mathrm{sinh}\:{x}\:=\frac{\mathrm{e}^{{x}} −\mathrm{e}^{−{x}} }{\mathrm{2}} \\ $$$$\mathrm{cos}\:{x}\:=\frac{\mathrm{e}^{\mathrm{i}{x}} +\mathrm{e}^{\mathrm{i}{x}} }{\mathrm{2}} \\ $$$$\mathrm{sin}\:{x}\:=\frac{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{2i}} \\ $$

Commented by bemath last updated on 26/Jun/20

oo thank you sir

$$\mathrm{oo}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Rio Michael last updated on 26/Jun/20

thank you′all

$$\mathrm{thank}\:\mathrm{you}'\mathrm{all} \\ $$

Commented by mr W last updated on 26/Jun/20

MJS sir:  i think cosh (ln x) and (1/2)(x+(1/x)) are  not exactly the same.  the domain of cosh (ln x) is x>0,  but the domain of (1/2)(x+(1/x)) is x<0  and x>0.  therefore (3/(2 cosh (ln x)+3)) has really  a global maximum (3/6).

$${MJS}\:{sir}: \\ $$$${i}\:{think}\:\mathrm{cosh}\:\left(\mathrm{ln}\:{x}\right)\:{and}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{{x}}\right)\:{are} \\ $$$${not}\:{exactly}\:{the}\:{same}. \\ $$$${the}\:{domain}\:{of}\:\mathrm{cosh}\:\left(\mathrm{ln}\:{x}\right)\:{is}\:{x}>\mathrm{0}, \\ $$$${but}\:{the}\:{domain}\:{of}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{{x}}\right)\:{is}\:{x}<\mathrm{0} \\ $$$${and}\:{x}>\mathrm{0}. \\ $$$${therefore}\:\frac{\mathrm{3}}{\mathrm{2}\:\mathrm{cosh}\:\left(\mathrm{ln}\:{x}\right)+\mathrm{3}}\:{has}\:{really} \\ $$$${a}\:{global}\:{maximum}\:\frac{\mathrm{3}}{\mathrm{6}}. \\ $$

Commented by MJS last updated on 26/Jun/20

I think cosh ln x is defined for x∈R\{0}  because ln (−∣x∣) =πi+ln ∣x∣  and e^(πi+ln ∣x∣) =−∣x∣ and e^(−(iπ+ln x)) =−(1/(∣x∣))  ⇒ cosh ln (−∣x∣) =−((x^2 +1)/(2∣x∣))

$$\mathrm{I}\:\mathrm{think}\:\mathrm{cosh}\:\mathrm{ln}\:{x}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$\mathrm{because}\:\mathrm{ln}\:\left(−\mid{x}\mid\right)\:=\pi\mathrm{i}+\mathrm{ln}\:\mid{x}\mid \\ $$$$\mathrm{and}\:\mathrm{e}^{\pi\mathrm{i}+\mathrm{ln}\:\mid{x}\mid} =−\mid{x}\mid\:\mathrm{and}\:\mathrm{e}^{−\left(\mathrm{i}\pi+\mathrm{ln}\:{x}\right)} =−\frac{\mathrm{1}}{\mid{x}\mid} \\ $$$$\Rightarrow\:\mathrm{cosh}\:\mathrm{ln}\:\left(−\mid{x}\mid\right)\:=−\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}\mid{x}\mid} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com