Question Number 199781 by cortano12 last updated on 09/Nov/23 $$\:\:\: \\ $$ Answered by qaz last updated on 09/Nov/23 $${f}\left({x}\right)=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\left(\frac{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}−{x}}\right)}{\frac{\mathrm{1}}{\mathrm{1}−{x}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)}−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{2}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)}{{x}}+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\…
Question Number 199167 by tri26112004 last updated on 28/Oct/23 $${Give}\:{a}\:{function}\: \\ $$$${f}:\:{R}\rightarrow\left(\mathrm{0};+\infty\right)\:{continous}\:{on}\:{R}\:{and}\:{such}\:{that} \\ $$$${f}\left({x}+{y}\right)\:=\:{f}\left({x}\right).{f}\left({y}\right) \\ $$$${a}.\:{Prove}\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$$${b}.\:{Let}\:{h}\left({x}\right)\:=\:{ln}\left[{f}\left({x}\right)\right].\:{Prove}\:{that}: \\ $$$$\:{h}\left({x}+{y}\right)\:=\:{h}\left({x}\right)\:+\:{h}\left({y}\right) \\ $$$${c}.\:{Find}\:{all}\:{the}\:{function}\:{f}\:{such}\:{that}\:{problem}\:{request} \\ $$$$\:\:\: \\…
Question Number 198541 by HomeAlone last updated on 21/Oct/23 $${s}\mathrm{1}={Abgdeeezdii}/{ii}\natural\gg{i} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 198279 by universe last updated on 16/Oct/23 $$\:\:\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{also}\:\mathrm{differentiable}\:\mathrm{on}\:\mathbb{R}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\mathrm{f}'\left(\mathrm{x}\right)\:>\:\mathrm{f}\left(\mathrm{x}\right)\:\forall\:\mathrm{x}\:\in\:\mathbb{R}\:{and}\:\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)\:=\:\mathrm{0}\:\mathrm{then}\: \\ $$$$\:\:\mathrm{prove}\:\mathrm{that}\:\:\mathrm{f}\left(\mathrm{x}\right)\:\geqslant\:\mathrm{0}\:\forall\:\mathrm{x}\:>\:\mathrm{x}_{\mathrm{0}} \\ $$ Answered by witcher3 last updated on 16/Oct/23 $$\mathrm{f}'\left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{x}\right)>\mathrm{0}….\left(\mathrm{1}\right)…
Question Number 198178 by universe last updated on 13/Oct/23 $${f}\left({xf}\left({y}\right)+{x}\right)={xy}+{f}\left({x}\right) \\ $$$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)=? \\ $$ Answered by Rasheed.Sindhi last updated on 13/Oct/23 $${Let}\:{f}\left({x}\right)={ax}+{b} \\…
Question Number 198064 by giovi last updated on 09/Oct/23 $$\mathrm{3}×\mathrm{5}^{{x}} +\mathrm{5}^{{x}+\mathrm{1}} =\mathrm{8}×\mathrm{5}^{\mathrm{3}} \\ $$ Answered by a.lgnaoui last updated on 09/Oct/23 $$\mathrm{3}×\mathrm{5}^{\mathrm{x}} +\mathrm{5}×\mathrm{5}^{\mathrm{x}} =\mathrm{8}×\mathrm{5}^{\mathrm{3}} \\…
Question Number 198065 by giovi last updated on 09/Oct/23 $$\mathrm{2}^{{x}} +\mathrm{9}+\mathrm{2}^{{x}} =\mathrm{40} \\ $$ Answered by a.lgnaoui last updated on 09/Oct/23 $$\:\mathrm{2}^{\mathrm{x}+\mathrm{1}} =\mathrm{31}\Rightarrow\:\:\left(\:\mathrm{x}+\mathrm{1}\right)\mathrm{ln2}=\mathrm{ln31} \\ $$$$\:\:\mathrm{x}=\frac{\mathrm{ln31}}{\mathrm{ln2}}−\mathrm{1}\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{ln}}\mathrm{31}−\boldsymbol{\mathrm{ln}}\mathrm{2}}{\boldsymbol{\mathrm{ln}}\mathrm{2}}…
Question Number 197972 by mnjuly1970 last updated on 07/Oct/23 $$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:{Find}\: \\ $$$$\:\:\:\:\:\:\:\mathscr{L}\:^{−\mathrm{1}} \left\{\:\:\frac{\mathrm{1}}{\mathrm{2}^{\:{s}} \:\sqrt{\:\mathrm{2}{s}+\mathrm{1}}\:}\:\right\}=\:? \\ $$$$ \\ $$$$\:\:\:\:\:\:{inverse}\:\:{laplace}\:{transform}… \\ $$ Terms of Service…
Question Number 197132 by Erico last updated on 08/Sep/23 $$\underset{\:\mathrm{0}} {\int}^{\:+\infty} \left(\frac{\mathrm{ln}\left(\mathrm{t}+\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)}{\mathrm{t}}\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$ Commented by mnjuly1970 last updated on 09/Sep/23 $$\:\:\:\:{ti}\:{tok}\:{agha}\:{sardool}\:..…
Question Number 196841 by Erico last updated on 01/Sep/23 $$\mathrm{Prove}\:\mathrm{that}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com