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

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