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Category: Relation and Functions

f-x-is-a-polynomial-such-that-f-x-2-f-x-f-x-1-Find-f-x-such-that-f-x-0-

Question Number 830 by prakash jain last updated on 20/Mar/15 $${f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{such}\:\mathrm{that} \\ $$$${f}\left({x}^{\mathrm{2}} \right)={f}\left({x}\right){f}\left({x}−\mathrm{1}\right) \\ $$$$\mathrm{Find}\:{f}\left({x}\right)\:\mathrm{such}\:\mathrm{that}\:{f}\left({x}\right)\neq\mathrm{0}. \\ $$ Commented by 123456 last updated on 23/Mar/15…

f-R-R-g-R-R-f-xy-f-x-g-x-f-y-g-y-g-xy-f-x-g-y-f-y-g-x-d-fg-dx-

Question Number 818 by 123456 last updated on 17/Mar/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({xy}\right)={f}\left({x}\right){g}\left({x}\right)+{f}\left({y}\right){g}\left({y}\right) \\ $$$${g}\left({xy}\right)={f}\left({x}\right){g}\left({y}\right)+{f}\left({y}\right){g}\left({x}\right) \\ $$$$\frac{{d}\left({fg}\right)}{{dx}}=? \\ $$ Commented by prakash jain last…

if-p-x-is-a-polynomial-and-p-x-p-1-x-p-x-p-1-x-p-3-28-then-p-x-p-4-

Question Number 806 by 123456 last updated on 16/Mar/15 $${if}\:{p}\left({x}\right)\:{is}\:{a}\:{polynomial}\:{and} \\ $$$${p}\left({x}\right){p}\left(\frac{\mathrm{1}}{{x}}\right)={p}\left({x}\right)+{p}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$${p}\left(\mathrm{3}\right)=\mathrm{28} \\ $$$${then} \\ $$$${p}\left({x}\right)=? \\ $$$${p}\left(\mathrm{4}\right)=? \\ $$ Commented by 123456…

let-f-x-cosx-1-x-1-prove-that-f-x-1-x-2-x-2-8-x-0-2-ptove-that-f-x-2-pi-e-1-x-1-ln-cosx-x-pi-2-

Question Number 66322 by mathmax by abdo last updated on 12/Aug/19 $${let}\:{f}\left({x}\right)=\left({cosx}\right)^{\frac{\mathrm{1}}{{x}}} \:\left(\:\mathrm{1}\right)\:\:{prove}\:{that}\:{f}\left({x}\right)\sim\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{8}}\:\:\left(\:{x}\rightarrow\mathrm{0}\right) \\ $$$$\left(\mathrm{2}\right){ptove}\:{that}\:{f}^{'} \left({x}\right)\sim−\frac{\mathrm{2}}{\pi}\:{e}^{\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right){ln}\left({cosx}\right)} \:\:\left({x}\rightarrow\frac{\pi}{\mathrm{2}}\right) \\ $$ Terms of Service Privacy Policy…