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Question Number 144270 by enter last updated on 24/Jun/21

Answered by ArielVyny last updated on 24/Jun/21

$$wehave\frac{d}{dx}\left(f\left(x\right)\right)⩾0because$$f'(x)-f(x)>=0 $$\frac{f\left(x\right)-f\left(1\right)}{x-1}=f^{\prime }\left(1\right)fiscontinuousin1andderivable$$$$f^{\prime }\left(1\right)-f\left(1\right)⩾0$$$$\frac{f\left(x\right)-f\left(1\right)}{x-1}-f\left(1\right)⩾0$$$$f\left(x\right)-f\left(1\right)⩾(x-1)f\left(1\right)$$$$f\left(x\right)⩾f\left(1\right)+(x-1)f\left(1\right)$$$$f\left(x\right)⩾xf\left(1\right)x⩾0andf\left(1\right)⩾0because{f}^{\prime }\left(x\right)⩾0andf\left(0\right)=0$$$$\mathit{c}\mathit{o}\mathit{n}\mathit{c}\mathit{l}\mathit{u}\mathit{s}\mathit{i}\mathit{o}\mathit{n}\mathit{f}\left(\mathit{x}\right)⩾0$$

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