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Question Number 214176 by a.lgnaoui last updated on 30/Nov/24
find the vslue of R(radius of circle betwen two    diagrame curves f(x)and g(x) as showen.  you can use the programer calculator.
$$\mathrm{find}\:\mathrm{the}\:\mathrm{vslue}\:\mathrm{of}\:\boldsymbol{\mathrm{R}}\left(\mathrm{radius}\:\boldsymbol{\mathrm{o}}\mathrm{f}\:\mathrm{circle}\:\mathrm{betwen}\:\mathrm{two}\:\right. \\ $$$$\:\mathrm{diagrame}\:\mathrm{curves}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\mathrm{and}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)\:\mathrm{as}\:\mathrm{showen}. \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{the}\:\mathrm{programer}\:\mathrm{calculator}. \\ $$
Commented by a.lgnaoui last updated on 30/Nov/24
Commented by mr W last updated on 30/Nov/24
impossible! f(x) is not symmetric!
$${impossible}!\:{f}\left({x}\right)\:{is}\:{not}\:{symmetric}! \\ $$
Commented by mr W last updated on 30/Nov/24
it′s possible if f(x)=(3/x^2 ) or  f(x)=(3/(∣x∣^3 ))+(1/(15∣x∣)).
$${it}'{s}\:{possible}\:{if}\:{f}\left({x}\right)=\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:{or} \\ $$$${f}\left({x}\right)=\frac{\mathrm{3}}{\mid{x}\mid^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{15}\mid{x}\mid}. \\ $$
Commented by a.lgnaoui last updated on 30/Nov/24
yes exactly .thank you
$$\mathrm{yes}\:\mathrm{exactly}\:.\mathrm{thank}\:\mathrm{you} \\ $$

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