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-

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

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

Commented by greougoury555 last updated on 11/Jun/22

$${y}'\:=\:\frac{\mathrm{1}.\left(\mathrm{1}+{x}\right)−\mathrm{1}.{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$$\:\begin{cases}{\mathrm{1}+{x}=\mathrm{1}\Rightarrow{x}=\mathrm{0}\:\wedge{y}=\mathrm{0}}\\{\mathrm{1}+{x}=−\mathrm{1}\Rightarrow{x}=−\mathrm{2}\:\wedge\:{y}=\:\mathrm{2}}\end{cases} \\ $$$$\:{eq}\:{of}\:{tangents}\:\begin{cases}{{y}={x}}\\{{y}={x}+\mathrm{4}}\end{cases} \\ $$

Commented by pete last updated on 11/Jun/22

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

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