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Category: Relation and Functions

solve-by-Laplace-transform-y-5y-2y-x-2-cosx-with-y-o-1-and-y-0-2-

Question Number 95212 by mathmax by abdo last updated on 24/May/20 $$\mathrm{solve}\:\mathrm{by}\:\mathrm{Laplace}\:\mathrm{transform} \\ $$$$\mathrm{y}^{''} \:+\mathrm{5y}^{'} \:+\mathrm{2y}\:=\mathrm{x}^{\mathrm{2}} \mathrm{cosx}\:\:\mathrm{with}\:\mathrm{y}\left(\mathrm{o}\right)=\mathrm{1}\:\mathrm{and}\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=\mathrm{2} \\ $$ Answered by mathmax by abdo…

how-do-you-solve-f-x-2-f-1-1-x-x-for-f-

Question Number 95150 by bobhans last updated on 23/May/20 $$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{f}\left(\mathrm{x}\right)\:+\mathrm{2}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\mathrm{x}\: \\ $$$$\mathrm{for}\:\mathrm{f}\:?\: \\ $$ Answered by mr W last updated on 23/May/20 $${f}\left({x}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)={x}\:\:\:…\left({i}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+\mathrm{2}{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:…\left({ii}\right)…

what-is-range-of-a-function-f-x-x-1-x-2-1-

Question Number 95140 by i jagooll last updated on 23/May/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{range}\:\mathrm{of}\:\mathrm{a}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{x}+\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}\: \\ $$ Commented by bobhans last updated on 23/May/20 $$\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{is}\:\mathrm{valid}\:\mathrm{when}\:\mathrm{x}^{\mathrm{2}}…

let-give-f-x-x-2-cos-1-x-2-if-x-0-1-but-its-derivative-f-is-not-integrable-on-0-1-

Question Number 29554 by abdo imad last updated on 09/Feb/18 $$\left.{l}\left.{et}\:{give}\:{f}\left({x}\right)=\:{x}^{\mathrm{2}} {cos}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:{if}\:{x}\in\right]\mathrm{0},\mathrm{1}\right]\:{but}\:{its}\:{derivative}\:{f}^{'} \\ $$$$\left.{i}\left.{s}\:{not}\:{integrable}\:{on}\:\right]\mathrm{0},\mathrm{1}\right]. \\ $$ Commented by abdo imad last updated on 14/Feb/18…