Question #312320

Find the equations of the tangents to the graph of y=x+1/x that are parallel to y+2x=0




1
Expert's answer
2022-03-18T15:15:23-0400

Slope of the line y+2x=0: m=-2.

Slope of the tangent: y=11x2.y'=1-\frac{1}{x^2}.

We have: y=my'=m or 11x2=2.1-\frac{1}{x^2}=-2.

So, x=13,  x=13.x=\frac{1}{\sqrt{3}},\;x=-\frac{1}{\sqrt{3}}.

y(13)=13+3.y(\frac{1}{\sqrt{3}})=\frac{1}{\sqrt{3}}+\sqrt{3}.

y(13)=133.y(-\frac{1}{\sqrt{3}})=-\frac{1}{\sqrt{3}}-\sqrt{3}.

Equations of the tangent lines:

y133=2(x13)y-\frac{1}{\sqrt{3}}-\sqrt{3}=-2(x-\frac{1}{\sqrt{3}}) or y=2x+23y=-2x+2\sqrt{3} .

y+13+3=2(x+13)y+\frac{1}{\sqrt{3}}+\sqrt{3}=-2(x+\frac{1}{\sqrt{3}}) or y=2x23.y=-2x-2\sqrt{3}.


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