Question Number 100388 by Ar Brandon last updated on 26/Jun/20 $$\:\:\:\:\:\:\:\mathcal{G}\mathrm{iven}\:\mathrm{f}:\left[\mathrm{0},\mathrm{2}\right]\rightarrow\mathbb{R}\:,\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{twice}\:\mathrm{derivable}\:\mathrm{and}\: \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{f}\left(\mathrm{1}\right)=\mathrm{f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$${i}-\mathcal{S}\mathrm{how}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{c}_{\mathrm{1}} ,\:\mathrm{c}_{\mathrm{2}} ,\:\mathrm{such}\:\mathrm{that}\:\mathrm{f}'\left(\mathrm{c}_{\mathrm{1}} \right)=\mathrm{0}\: \\ $$$$\mathrm{and}\:\mathrm{f}'\left(\mathrm{c}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$${ii}-\mathcal{S}\mathrm{how}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{c}_{\mathrm{3}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{f}''\left(\mathrm{c}_{\mathrm{3}} \right)=\mathrm{0}…
Question Number 100391 by Ar Brandon last updated on 26/Jun/20 $$\:\:\:\mathcal{D}\mathrm{etermine}\:\mathrm{the}\:\mathrm{coordinates}\:\mathrm{where}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c} \\ $$$$\mathrm{admits}\:\mathrm{a}\:\mathrm{local}\:\mathrm{point}. \\ $$ Answered by MJS last updated on 26/Jun/20 $${ax}^{\mathrm{2}} +{bx}+{c}={x}…
Question Number 34821 by ajfour last updated on 11/May/18 $${Find}\:{range}\:{of} \\ $$$$\:\:\:{y}=\frac{{x}}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)}\:. \\ $$ Commented by ajfour last updated on 11/May/18 Answered by ajfour last…
Question Number 165849 by daus last updated on 09/Feb/22 $${solve}\:{the}\:{differential}\:{equation} \\ $$$$\frac{{dy}}{{dx}}+\frac{{y}}{{x}−\mathrm{1}}=\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$ Commented by mkam last updated on 10/Feb/22 $$\boldsymbol{{p}}\left(\boldsymbol{{x}}\right)=\:\frac{\mathrm{1}}{\boldsymbol{{x}}−\mathrm{1}}\:\:,\:\boldsymbol{{Q}}\left(\boldsymbol{{x}}\right)\:=\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\mathrm{1}} \\ $$$$ \\…
Question Number 100243 by mhmd last updated on 25/Jun/20 $${find}\:{laplace}\:{transforme}\:{of}\:{the}\:{function}\: \\ $$$${f}\left({t}\right)=\left({a}−{bt}\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \left({wt}\right)? \\ $$$$ \\ $$$${help}\:{me}\:{sir}\:? \\ $$ Answered by mathmax by abdo…
Question Number 165746 by Eric002 last updated on 07/Feb/22 $${compute}\:{the}\:{extreme}\:{points}\:{of}:\: \\ $$$${f}={e}^{{x}} {sin}\left({x}+{y}\right) \\ $$ Answered by TheSupreme last updated on 07/Feb/22 $${y}=\mathrm{0} \\ $$$${f}\left({x},\mathrm{0}\right)={e}^{{x}}…
Question Number 165687 by mnjuly1970 last updated on 06/Feb/22 $$ \\ $$$$\:\:\mathrm{I}{f}\:\:\:\:{f}\left({x}\right)=\:\frac{{x}^{\:\mathrm{2}} −\:\mathrm{2}{x}\:−\mathrm{8}}{{x}^{\:\mathrm{2}} −\mathrm{7}{x}\:+\mathrm{12}} \\ $$$$\:\:\:{then}\:,{find}\::\:\:\:\:\:\:\:\:\:\:\:{f}^{\:−\mathrm{1}} \left({x}\right)=? \\ $$$$ \\ $$ Answered by TheSupreme last…
Question Number 165641 by nadovic last updated on 05/Feb/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Given}\:\mathrm{that}\:\:{y}\:=\:\frac{\mathrm{1}}{{x}}\: \\ $$$$\left({a}\right)\:\mathrm{Show}\:\mathrm{that}\:\:{y}^{\left({n}\right)} \:=\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{{x}^{{n}+\mathrm{1}} } \\ $$$$\left({b}\right)\:\mathrm{Find}\:\mathrm{an}\:\mathrm{expression}\:\mathrm{for}\:{y}^{\left({n}−\mathrm{1}\right)} +\:{y}^{\left({n}\right)} \\ $$$$ \\ $$ Answered by aleks041103…
Question Number 165597 by mnjuly1970 last updated on 05/Feb/22 $$ \\ $$$$\:\:\:\:{prove}\:{that} \\ $$$$\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\psi^{\:\left(\mathrm{1}\right)} \left({n}\right)}{{n}^{\:\mathrm{2}} }\:=\frac{\mathrm{7}}{\mathrm{4}}\:\zeta\:\left(\mathrm{4}\right)\:\:\:\blacksquare\:{m}.{n} \\ $$$$ \\ $$ Terms of Service…
Question Number 165599 by mnjuly1970 last updated on 05/Feb/22 Answered by MJS_new last updated on 05/Feb/22 $${f}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:−\mathrm{1}<{x}<\mathrm{1}\:\Rightarrow\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}<{f}\left({x}\right)\leqslant\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact:…