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Question-125639

Question Number 125639 by aurpeyz last updated on 12/Dec/20 Commented by ZiYangLee last updated on 13/Dec/20 $$\mathrm{A}\:\mathrm{more}\:\mathrm{alternative}\:\mathrm{ways}: \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}}−\mathrm{1} \\ $$$$\Rightarrow{f}'\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty}…

Consider-a-continuously-differentiable-function-f-0-1-R-such-that-f-0-0-and-f-1-1-Find-the-minimum-value-of-0-1-f-x-2-1-x-2-dx-

Question Number 125621 by ZiYangLee last updated on 12/Dec/20 $$\mathrm{Consider}\:\mathrm{a}\:\mathrm{continuously}\:\mathrm{differentiable} \\ $$$$\mathrm{function}\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R}\:\mathrm{such}\:\mathrm{that}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{and}\:{f}\left(\mathrm{1}\right)=\mathrm{1}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({f}'\left({x}\right)\right)^{\mathrm{2}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$ Terms of Service…

f-x-f-1-x-x-2-gt-f-x-

Question Number 191151 by TUN last updated on 19/Apr/23 $${f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)={x}^{\mathrm{2}} \\ $$$$=>{f}\left({x}\right)=¿ \\ $$ Answered by Rasheed.Sindhi last updated on 19/Apr/23 $${f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)={x}^{\mathrm{2}} …….\left({i}\right) \\ $$$${Replacing}\:{x}\:{by}\:\mathrm{1}−{x}:…

Question-191131

Question Number 191131 by TUN last updated on 18/Apr/23 Answered by a.lgnaoui last updated on 19/Apr/23 $$\bigtriangleup=\left(\mathrm{2m}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{m}^{\mathrm{2}} −\mathrm{6}\right)=\mathrm{25}−\mathrm{4m} \\ $$$$\bigtriangleup>\mathrm{0}\:\:\:\Rightarrow\mathrm{pour}\:\:\:\:\mathrm{m}<\frac{\mathrm{25}}{\mathrm{4}}\:\:\:\:\:\:\mathrm{2}\:\mathrm{solutions} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{2m}−\mathrm{1}−\sqrt{\mathrm{25}−\mathrm{4m}}}{\mathrm{2}};\:\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{2m}−\mathrm{1}+\sqrt{\mathrm{25}−\mathrm{4m}}}{\mathrm{2}}…

Show-that-in-a-30-60-90-triangle-the-altitude-on-the-hypotaneuse-divides-the-hypotaneuse-into-segments-whose-length-has-the-ratio-1-3-without-using-trigonometry-

Question Number 60043 by Kunal12588 last updated on 17/May/19 $${Show}\:{that}\:{in}\:{a}\:\mathrm{30}°−\mathrm{60}°−\mathrm{90}°\:{triangle}\:{the}\: \\ $$$${altitude}\:{on}\:{the}\:{hypotaneuse}\:{divides}\:{the}\: \\ $$$${hypotaneuse}\:{into}\:{segments}\:{whose}\:{length} \\ $$$${has}\:{the}\:{ratio}\:\mathrm{1}/\mathrm{3}. \\ $$$${without}\:{using}\:{trigonometry}. \\ $$ Answered by tanmay last updated…