Question Number 131734 by LYKA last updated on 07/Feb/21 $$\boldsymbol{{given}}\:\boldsymbol{{the}}\:\boldsymbol{{function}} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}.\boldsymbol{{y}}\right)=\boldsymbol{{xy}}\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{y}}−\mathrm{1}\right) \\ $$$$\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}.\boldsymbol{{y}}\right)\:\boldsymbol{{has}}\:\boldsymbol{{some}}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\boldsymbol{{as}}\:\boldsymbol{{a}}\:\boldsymbol{{stationery}}\:\boldsymbol{{point}} \\ $$$$ \\ $$$$\boldsymbol{{use}}\:\boldsymbol{{tylor}}\:\boldsymbol{{series}}\:\boldsymbol{{method}}\:\boldsymbol{{to}}\: \\ $$$$\boldsymbol{{determine}}\:\boldsymbol{{whether}}\:\left(\mathrm{0}.\mathrm{1}\right)\:\boldsymbol{{is}}\:\boldsymbol{{a}} \\ $$$$\boldsymbol{{minima}}\:,\boldsymbol{{maxima}}\:\boldsymbol{{or}}\:\boldsymbol{{saddle}}\: \\…
Question Number 66197 by Rasheed.Sindhi last updated on 11/Aug/19 $$\mathrm{If}\:\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{1},\mathrm{prove}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{n}} +\mathrm{x}^{\mathrm{n}−\mathrm{2}} +\mathrm{x}^{\mathrm{n}−\mathrm{4}} =\mathrm{0} \\ $$ Commented by mr W last updated on 10/Aug/19…
Question Number 131729 by ajfour last updated on 07/Feb/21 Commented by ajfour last updated on 07/Feb/21 $${Find}\:{radius}\:{of}\:{sphere}\:{of}\:{density} \\ $$$$\rho\:{that}\:{receives}\:{maximum} \\ $$$${normal}\:{reaction}\:{from}\:{wall}. \\ $$ Answered by…
Question Number 66194 by aliesam last updated on 10/Aug/19 $$\int\frac{{x}^{{a}} }{{bx}^{{n}} +{c}}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 658 by 123456 last updated on 22/Feb/15 $${proof}\:{that}\:{n}!>\left(\frac{{n}}{\mathrm{3}}\right)^{{n}} ,{n}\in\mathbb{N}^{\ast} \\ $$ Commented by 123456 last updated on 20/Feb/15 $${n}=\mathrm{1}\Rightarrow\mathrm{1}!=\mathrm{1}>\frac{\mathrm{1}}{\mathrm{3}}=\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{1}} \\ $$$${n}=\mathrm{1}\Rightarrow\mathrm{0}!=\mathrm{1}>\frac{\mathrm{1}}{\mathrm{3}}\approx\mathrm{0}.\mathrm{33} \\ $$$${n}=\mathrm{2}\Rightarrow\mathrm{2}!=\mathrm{2}>\frac{\mathrm{4}}{\mathrm{9}}=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}}…
Question Number 657 by 123456 last updated on 21/Feb/15 $${if}\:\left({a}_{{n}} \right)\:{and}\:\left({b}_{{n}} \right)\:{are}\:{two}\:{real}\:{sequence} \\ $$$${such}\:{that}\:{e}^{{a}_{{n}} } ={a}_{{n}} +{e}^{{b}_{{n}} } \\ $$$$\left.{a}\right)\:{proof}\:{that}\:{a}_{{n}} >\mathrm{0}\Rightarrow{b}_{{n}} >\mathrm{0} \\ $$$$\left.{b}\right)\:{if}\:{a}_{{n}} >\mathrm{0}\forall{n}\in\mathbb{N}\:{if}\:\underset{{n}=\mathrm{0}}…
Question Number 655 by 123456 last updated on 19/Feb/15 $${f}\left({z}\right)=\frac{\mathrm{1}−\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{2}+{z}^{\mathrm{2}} }}{\mathrm{2}+\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{2}+{z}^{\mathrm{2}} }} \\ $$$$\frac{{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)}{\mathrm{2}}−{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=? \\ $$ Answered by prakash jain last updated on…
Question Number 652 by 123456 last updated on 19/Feb/15 $${if}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:{is}\:{continuous}\:{and} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right)+{y} \\ $$$$\mathrm{1}.\:{find}\:{f}\left({x}\right) \\ $$$$\mathrm{2}.\:{proof}\:{or}\:{disproof}\:{that}\:{f}'\left({x}\right)=\mathrm{1} \\ $$$$\mathrm{3}.\:{if}\:{f}\left(\mathrm{0}\right)=\mathrm{0}\:{proof}\:{or}\:{disproof}\:{that}\:{f}\left({x}\right)={x} \\ $$ Answered by prakash jain last…
Question Number 650 by 123456 last updated on 19/Feb/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right){g}\left({y}\right) \\ $$$${g}\left({x}+{y}\right)={f}\left({x}\right){g}\left({x}\right)+{g}\left({y}\right) \\ $$ Answered by prakash jain last updated on…
Question Number 66185 by Tony Lin last updated on 10/Aug/19 $$\left({e}^{\frac{\mathrm{1}}{{e}}} \right)^{\left({e}^{\frac{\mathrm{1}}{{e}}} \right)^{.\centerdot^{.\left({e}^{\frac{\mathrm{1}}{{e}}} \right)} } } =? \\ $$ Commented by GordonYeeman last updated on…