Question Number 187053 by horsebrand11 last updated on 13/Feb/23 $$\:{How}\:{do}\:{you}\:{make}\:{a}\:{curve}\: \\ $$$$\:{y}={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}\:{with}\:{a}\:{critical} \\ $$$$\:{point}\:{of}\:\left(\mathrm{1},\mathrm{0}\right)\:{and}\:\left(−\mathrm{2},\mathrm{27}\right)\:? \\ $$ Answered by cortano12 last updated on 13/Feb/23…
Question Number 121400 by rs4089 last updated on 07/Nov/20 $${ind}\:{maximum}\:{value}\:{of}\:\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} } \\ $$ Answered by MJS_new last updated on 07/Nov/20 $$\mathrm{for}\:{x},\:{y}\:\in\mathbb{R}:\:\mathrm{0}\leqslant\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }<+\infty \\…
Question Number 121368 by abdullahquwatan last updated on 07/Nov/20 $$\mathrm{find}\:\mathrm{minimun}\:\mathrm{and}\:\mathrm{maksimum} \\ $$$${y}=\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{3}} \\ $$ Answered by MJS_new last updated on 07/Nov/20 $${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\:\mathrm{zero}\:\mathrm{at}\:{x}=\mathrm{1}…
Question Number 121361 by rs4089 last updated on 07/Nov/20 Answered by TANMAY PANACEA last updated on 07/Nov/20 $${u}={x}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)−{y}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\frac{{x}}{{y}}\right) \\ $$$$={x}^{\mathrm{3}} \left\{{sin}^{−\mathrm{1}}…
Question Number 121362 by mnjuly1970 last updated on 07/Nov/20 $$\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:{evaluate}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{K}=\underset{{n}\in\mathbb{N}} {\sum}\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\right)}\:=? \\ $$ Commented by mindispower last updated on…
Question Number 121353 by john santu last updated on 06/Nov/20 Answered by liberty last updated on 07/Nov/20 $$\mathrm{Let}\:\mathrm{x}\:\mathrm{be}\:\mathrm{the}\:\mathrm{x}−\mathrm{coordinate}\:\mathrm{of}\:\mathrm{end}\:\mathrm{point}\: \\ $$$$\mathrm{that}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{x}−\mathrm{axis}\:\mathrm{and}\:\mathrm{let}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{endpoint}\:\mathrm{be}\:\left(\mathrm{0},\mathrm{y}\right).\:\mathrm{Thus}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{function} \\ $$$$\mathrm{f}\:\mathrm{which}\:\mathrm{gives}\:\mathrm{y}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{x}.\:\mathrm{Since} \\…
Question Number 121282 by mnjuly1970 last updated on 06/Nov/20 $$\:\:\:\:\:\:\:\:…\:\mathrm{advanced}\:\:\mathrm{mathematics}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{prove}\:\:\mathrm{that}\:\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Psi=\int_{\mathrm{0}} ^{\:\mathrm{1}} \Gamma\left(\mathrm{2}−{x}\right)\Gamma\left(\mathrm{1}+{x}\right){dx}=\frac{\mathrm{7}}{\pi^{\mathrm{2}} }\:\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\mathrm{m}.\mathrm{n}.\mathrm{july}.\mathrm{1970}… \\ $$$$ \\ $$…
Question Number 121264 by benjo_mathlover last updated on 06/Nov/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{x}=\mathrm{2y}^{\mathrm{2}} \:\mathrm{closest}\:\mathrm{to}\:\mathrm{the}\:\mathrm{point} \\ $$$$\left(\mathrm{10},\mathrm{0}\right) \\ $$ Answered by liberty last updated on 06/Nov/20 $$\mathrm{Let}\:\mathrm{L}\:\mathrm{be}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{distance}\:\mathrm{between}…
Question Number 186752 by depressiveshrek last updated on 09/Feb/23 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{7}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{4}} +\mathrm{3}}}{\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} +\mathrm{2}{x}^{\mathrm{5}} +\mathrm{1}}−\sqrt[{\mathrm{5}}]{{x}^{\mathrm{7}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}}} \\ $$$${Please}\:{show}\:{work}. \\ $$ Answered by Ar…
Question Number 121202 by benjo_mathlover last updated on 05/Nov/20 Commented by liberty last updated on 05/Nov/20 $$\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{x}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:>\:\mathrm{0}\:,\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{increasing}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\mathrm{interval}\:\left(−\infty,\mathrm{0}\right)\:\cup\left(\mathrm{0},\infty\right)\:.\:\mathrm{Then}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{no}\:\mathrm{have}\:\mathrm{local}\:\mathrm{maxima}\:\mathrm{and}\:\mathrm{local}\:\mathrm{minima}…