Find-the-coordinates-of-the-point-on-the-curve-y-x-1-x-at-which-the-tangents-to-the-curve-are-parallel-to-the-line-x-y-8-0-Find-the-equations-of-the-tangents-at-these-points-

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Question Number 171267 by pete last updated on 11/Jun/22

$$\mathrm{Find}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{curve}$$$$\mathrm{y}=\frac{x}{1+x}\mathrm{at}\mathrm{which}\mathrm{the}\mathrm{tangents}\mathrm{to}\mathrm{the}\mathrm{curve}$$$$\mathrm{are}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}x-y+8=0.$$$$\mathrm{Find}\mathrm{the}\mathrm{equations}\mathrm{of}\mathrm{the}\mathrm{tangents}\mathrm{at}$$$$\mathrm{these}\mathrm{points}.$$

Commented by greougoury555 last updated on 11/Jun/22

$$y^{\prime }=\frac{1.(1+x)-1.x}{{(1+x)}^2}=\frac{1}{{(1+x)}^2}=1$$$$\{\begin{array}{l}1+x=1\Rightarrow x=0\wedge y=0\\ 1+x=-1\Rightarrow x=-2\wedge y=2\end{array}$$$$eqoftangents\{\begin{array}{l}y=x\\ y=x+4\end{array}$$

Commented by pete last updated on 11/Jun/22

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

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