Category: Relation and Functions

Question Number 206442 by universe last updated on 14/Apr/24 Answered by Berbere last updated on 14/Apr/24 $$letf\left(x\right)=e^{g\left(x\right)};particularSolutionjusttosimplifie$$$$theproblems;f\left(0\right)=1=e^{g\left(0\right)}\Rightarrow g\left(0\right)=0$$$$\Rightarrow{f}'\left({x}\right)={g}'{e}^{{g}\left({x}\right)} ;{f}''\left({x}\right)=\left({g}'^{\mathrm{2}} +{g}''\right){e}^{{g}\left({x}\right)}…

Question Number 206136 by cortano21 last updated on 07/Apr/24 Answered by A5T last updated on 07/Apr/24 $$f\left(f\left(f\left(f\left(x\right)\right)\right)\right)={\left[f\left(x\right)\right]}^2-3\left[f\left(x\right)\right]+4=f(x^2-3x+4)$$$${\left[f\left(2\right)\right]}^2-3\left[f\left(2\right)\right]+4=f\left(2\right)\Rightarrow {\left[f\left(2\right)\right]}^2-4f\left(2\right)+4=0$$$$\left[{f}\left(\mathrm{2}\right)−\mathrm{2}\right]^{\mathrm{2}}…

Question Number 205262 by mathzup last updated on 13/Mar/24 $$natureoftheserie\sum _{n⩾1}\frac{ln\left(n\right)}{n}$$ Answered by Berbere last updated on 13/Mar/24 $$\mathrm{\forall }n⩾2ln\left(n\right)⩾ln\left(2\right)>\frac{1}{2}$$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{ln}\left({n}\right)}{{n}}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}\rightarrow+\infty\:{serie}\:{dv}…

Question Number 204426 by universe last updated on 17/Feb/24 $$\mathrm{let}\mathrm{a},\mathrm{b}>0\mathrm{find}\mathrm{all}\mathrm{differentiable}\mathrm{function}$$$$\mathrm{f}:(0,\mathrm{\infty })\to (0,\mathrm{\infty })\mathrm{such}\mathrm{that}$$$${\mathrm{f}}^{\prime }\left(\frac{a}{\mathrm{x}}\right)=\frac{\mathrm{bx}}{\mathrm{f}\left(\mathrm{x}\right)},\mathrm{\forall }\mathrm{x}>0$$ Answered by MrGaster last updated on 03/Feb/25 $$\mathrm{Let}\:{f}\left({x}\right)=\sqrt{{bx}^{\mathrm{2}} +{c}},{c}=\frac{{a}^{\mathrm{2}}…

Question Number 204393 by ks last updated on 15/Feb/24 Terms of Service Privacy Policy Contact: info@tinkutara.com