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Question Number 71816 by psyche last updated on 20/Oct/19

$$supposethatfiscontinuousanddifferentiablein(a,b)if{f}^{\prime }\left(x\right)=0,\mathrm{\forall }x\in (a,b)thenshowthatfisconstanton[a,b].$$

Answered by mind is power last updated on 20/Oct/19

$$\mathrm{assum}\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{consrante}$$$$\mathrm{\exists }\mathrm{x},\mathrm{y}\mathrm{suche}\mathrm{that}1⩾\mathrm{y}\ne \mathrm{x}⩾0\mathrm{f}\left(\mathrm{x}\right)\ne \mathrm{f}\left(\mathrm{y}\right)$$$$\mathrm{use}\mathrm{mean}\mathrm{value}\mathrm{theorem}\mathrm{for}\mathrm{f}\mathrm{in}[\mathrm{x},\mathrm{y}]\Rightarrow$$$$\mathrm{\exists }\mathrm{c}\in [\mathrm{x},\mathrm{y}]\mathrm{such}\mathrm{that}$$$$\frac{\mathrm{f}\left(\mathrm{y}\right)-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{y}-\mathrm{x}}={\mathrm{f}}^{\prime }\left(\mathrm{c}\right)\Rightarrow \mathrm{f}\left(\mathrm{y}\right)-\mathrm{f}\left(\mathrm{x}\right)=(\mathrm{y}-\mathrm{x}){\mathrm{f}}^{\prime }\left(\mathrm{c}\right)$$$$\mathrm{but}{\mathrm{f}}^{\prime }\left(\mathrm{c}\right)=0\mathrm{cause}\mathrm{c}\in [0,1]$$$$\Rightarrow \mathrm{f}\left(\mathrm{y}\right)-\mathrm{f}\left(\mathrm{x}\right)=0\Rightarrow \mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{absurd}\Rightarrow \mathrm{\forall }(\mathrm{x},\mathrm{y})\in {[0,1]}^2\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)$$$$\mathrm{f}\mathrm{constant}$$

Commented by psyche last updated on 24/Oct/19

$$thanks$$

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