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Question Number 80180 by mathocean1 last updated on 31/Jan/20

$$\mathrm{h}\left(x\right)=\frac{x-x^2}{x+1}$$$$wedefinedthisfunctionon$$$$\mathbb{R}-\{-1\}\to \mathbb{R}$$$$$$$$1)\mathrm{Study}\mathrm{the}\mathrm{variations}\mathrm{of}\mathrm{h}\mathrm{then}$$$$\mathrm{draw}\mathrm{up}\mathrm{its}\mathrm{table}\mathrm{of}\mathrm{variation}.$$$$$$$$\mathrm{please}\mathrm{sirs}\mathrm{i}\mathrm{need}\mathrm{your}\mathrm{kind}\mathrm{help}$$

Answered by Rio Michael last updated on 31/Jan/20

$$\mathbf{solution}$$$$----------$$$$\mathrm{h}\left(x\right)=\frac{x-x^2}{x-1}$$$$h\left(x\right)=0\Rightarrow x-x^2=0\Rightarrow x=0orx=1$$$${\mathrm{h}}^{\prime }\left(x\right)=\frac{(x-x^2)-(x-1)(1-2x)}{{(x-1)}^2}=\frac{x-x^2-(x-2x^2-1+2x)}{{(x-1)}^2}$$$$h^{{}^{\prime }}\left(x\right)=\frac{x^2-2x+1}{{(x-1)}^2}$$$$h^{{}^{\prime }}\left(x\right)>0\Rightarrow {(x-1)}^2>0\Rightarrow x>1$$$$hisincreasingforx>1anddecreasingotherwise$$$$$$

Commented by mathocean1 last updated on 31/Jan/20

$$\mathrm{Please}\mathrm{sir}{\mathrm{h}}^{\prime }\mathrm{is}\mathrm{not}\mathrm{like}\mathrm{this}:$$$${\mathrm{h}}^{\prime }=\frac{(1-2x)(x+1)-(x-x^2)}{{(x+1)}^2}?$$$$$$

Commented by mathocean1 last updated on 31/Jan/20

$$\mathrm{h}\left(x\right)=\frac{x-x^2}{x+1}$$

Commented by Rio Michael last updated on 31/Jan/20

$$sorryicopiedthequestionwrongly$$

Commented by mathocean1 last updated on 31/Jan/20

$$\mathrm{please}\mathrm{sir}\mathrm{how}\mathrm{can}\mathrm{i}\mathrm{draw}\mathrm{up}\mathrm{its}$$$$\mathrm{table}\mathrm{of}\mathrm{variation}\mathrm{when}$$$${\mathrm{h}}^{\prime }=\frac{-x^2-2x+1}{{(x+1)}^2}=-1?$$

Commented by JDamian last updated on 01/Feb/20

$$if\mathrm{h}\left(x\right)were\frac{x-x^2}{x-1},then\mathrm{h}\left(x\right)=\frac{x(1-x)}{x-1}=-x$$$$averyeasyfunction$$

Commented by mathocean1 last updated on 01/Feb/20

$$\mathrm{so}...\mathrm{we}\mathrm{can}\mathrm{study}\mathrm{h}\left(x\right)=\frac{x-x^2}{x+1}???$$

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