Question Number 140073 by EDWIN88 last updated on 04/May/21 $$\mathrm{Let}\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{3x}^{\mathrm{2}} −\mathrm{1}\:;\:\mathrm{x}<\mathrm{0}}\\{\mathrm{cx}+\mathrm{d}\:;\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1}}\\{\sqrt{\mathrm{x}+\mathrm{8}}\:;\:\mathrm{x}>\mathrm{1}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{c}\:\&\:\mathrm{d}\:\mathrm{such}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{continous} \\ $$$$\mathrm{everywhere} \\ $$ Answered by bobhans last updated on 04/May/21 $$\mathrm{since}\:\underset{{x}\rightarrow\mathrm{0}^{\:−}…
For-what-value-of-k-is-the-following-continous-function-f-x-7x-2-6x-4-x-2-if-x-2-7-amp-x-2-k-if-x-2-
Question Number 140071 by EDWIN88 last updated on 04/May/21 $$\mathrm{For}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{is}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{continous}\:\mathrm{function}\:? \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\frac{\sqrt{\mathrm{7x}+\mathrm{2}}−\sqrt{\mathrm{6x}+\mathrm{4}}}{\mathrm{x}−\mathrm{2}}\:;\:\mathrm{if}\:\mathrm{x}\geqslant−\frac{\mathrm{2}}{\mathrm{7}}\:\&\:\mathrm{x}\neq\mathrm{2}}\\{\:\:\:\:\:\:\:\:\mathrm{k}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:;\:\mathrm{if}\:\mathrm{x}=\mathrm{2}}\end{cases} \\ $$ Answered by bobhans last updated on 04/May/21 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow\mathrm{2}}…
Question Number 74501 by mathmax by abdo last updated on 25/Nov/19 $$\:{let}\:{P}\left({x}\right)=\frac{\mathrm{1}}{{n}!}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} \\ $$$${calculate}\:{P}^{\left({n}\right)} \left({x}\right)\:\:{and}\:{P}^{\:\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Commented by mathmax by…
Question Number 74498 by mathmax by abdo last updated on 25/Nov/19 $$\left.\mathrm{1}\right)\:{calculte}\:\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \left[{e}^{{x}} \right]\:{dx}\:\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{2}\right){find}\:{lim}_{{n}\rightarrow+\infty} \:{n}^{{n}} \:{A}_{{n}} \\ $$ Commented by…
Question Number 74499 by mathmax by abdo last updated on 25/Nov/19 $${decompose}\:{inside}\:{C}\left({x}\right)\:{the}\:{fraction} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} } \\ $$ Commented by mathmax by abdo last updated…
Question Number 139954 by Ar Brandon last updated on 02/May/21 Answered by mr W last updated on 03/May/21 $${x}_{{n}+\mathrm{1}} =\frac{{x}_{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}}{\mathrm{1}+{x}_{{n}} ×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}} \\ $$$${let}\:\mathrm{tan}\:{A}_{{n}} ={x}_{{n}}…
Question Number 74353 by mathmax by abdo last updated on 22/Nov/19 $${let}\:\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}+\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\left.\mathrm{1}\right){find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{a}\:{equivalent}\:{of}\:{A}_{{n}} \:\:{when}\:{n}\rightarrow+\infty \\ $$$$ \\…
Question Number 74352 by mathmax by abdo last updated on 22/Nov/19 $${let}\:{U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\:\:{find}\:{a}\:{equivalent}\:{of}\:{U}_{{n}} \:\:\:\left({n}\rightarrow+\infty\right) \\ $$$$ \\ $$ Answered by mind is…
Question Number 74350 by mathmax by abdo last updated on 22/Nov/19 $${findf}\left({a}\right)=\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({cosx}\right)}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}\:{witha}>\mathrm{0} \\ $$ Commented by abdomathmax last updated on 23/Nov/19…
Question Number 74351 by mathmax by abdo last updated on 22/Nov/19 $${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by ~blr237~ last updated on…