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Question Number 198279 by universe last updated on 16/Oct/23

$$\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{also}\mathrm{differentiable}\mathrm{on}\mathbb{R}\mathrm{such}\mathrm{that}$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)>\mathrm{f}\left(\mathrm{x}\right)\mathrm{\forall }\mathrm{x}\in \mathbb{R}and\mathrm{f}\left({\mathrm{x}}_0\right)=0\mathrm{then}$$$$\mathrm{prove}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)⩾0\mathrm{\forall }\mathrm{x}>{\mathrm{x}}_0$$

Answered by witcher3 last updated on 16/Oct/23

$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)>0....\left(1\right)$$$$\left(1\right)\ast {\mathrm{e}}^{-\mathrm{x}}\iff$$$$\mathrm{f}{\left(\mathrm{x}\right)}^{\prime }{\mathrm{e}}^{-\mathrm{x}}-\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{x}}>0$$$${\left(\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{x}}\right)}^{\prime }>0$$$$\Rightarrow {\int }_{{\mathrm{x}}_0}^{\mathrm{x}}{\left(\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{x}}\right)}^{\prime }\mathrm{dx}>{\int }_{{\mathrm{x}}_0}^{\mathrm{x}}0\mathrm{dx}$$$$\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{x}}-\mathrm{f}\left({\mathrm{x}}_0\right){\mathrm{e}}_{=0}^{-{\mathrm{x}}_0}⩾0$$$$\mathrm{f}\left(\mathrm{x}\right){\mathrm{e}}^{-\mathrm{x}}⩾0$$$$\mathrm{\forall }\mathrm{x}>0,\Rightarrow \mathrm{f}\left(\mathrm{x}\right)⩾0,\mathrm{x}\to {\mathrm{e}}^{-\mathrm{x}}>0,\mathrm{\forall }\mathrm{x}\in [{\mathrm{x}}_0,\mathrm{\infty }[$$$$$$$$$$

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