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Category: Relation and Functions

Question-136693

Question Number 136693 by liberty last updated on 25/Mar/21 Commented by Olaf last updated on 25/Mar/21 $${f}\left({x}\right)\:=\:{mx}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +{kx}−\mathrm{18} \\ $$$$=\:\left(\mathrm{6}−{x}−{x}^{\mathrm{2}} \right)\left(−{mx}−\mathrm{3}\right) \\ $$$$=\left({x}^{\mathrm{2}} +{x}−\mathrm{6}\right)\left({mx}+\mathrm{3}\right)…

f-x-x-x-1-x-2-x-10-f-0-

Question Number 70891 by Henri Boucatchou last updated on 09/Oct/19 $$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:=\:\boldsymbol{{x}}\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)….\left(\boldsymbol{{x}}−\mathrm{10}\right) \\ $$$$\boldsymbol{{f}}\:'\left(\mathrm{0}\right)\:=\:? \\ $$ Commented by kaivan.ahmadi last updated on 09/Oct/19 $${let}\:{g}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)…\left({x}−\mathrm{10}\right)\Rightarrow{f}\left({x}\right)={xg}\left({x}\right) \\ $$$$\Rightarrow{f}'\left({x}\right)={g}\left({x}\right)+{xg}'\left({x}\right)\Rightarrow…

letf-x-x-3-arctan-pi-x-1-calculate-f-n-x-2-calculate-f-n-1-3-developp-f-at-integer-serie-

Question Number 136370 by mathmax by abdo last updated on 21/Mar/21 $$\mathrm{letf}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} \:\mathrm{arctan}\left(\frac{\pi}{\mathrm{x}}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integer}\:\mathrm{serie} \\ $$ Answered by…

find-the-domain-and-range-of-y-4x-2-3-x-lt-2-3x-1-x-2-

Question Number 5237 by sanusihammed last updated on 02/May/16 $${find}\:{the}\:{domain}\:{and}\:{range}\:{of}\: \\ $$$${y}\:=\:\left[_{\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:<\mathrm{2}} ^{\:\mathrm{3}{x}−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\geqslant\:\mathrm{2}} \right. \\ $$ Answered by FilupSmith last updated on 02/May/16 $${Domain}\:=\:\left\{{x}:\:−\infty\leqslant{x}\leqslant\infty\right\}…