SOLUTION: Find the equation of the tangent to the curve defined by y=x-(e^-x) that is parallel to the line represented by 3x-y-9=0.

Algebra ->  Vectors -> SOLUTION: Find the equation of the tangent to the curve defined by y=x-(e^-x) that is parallel to the line represented by 3x-y-9=0.       Log On


   

Question 199658: Find the equation of the tangent to the curve defined by y=x-(e^-x) that is parallel to the line represented by 3x-y-9=0.
Answer by Alan3354(69443) About Me  (Show Source):
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Find the equation of the tangent to the curve defined by y=x-(e^-x) that is parallel to the line represented by 3x-y-9=0.
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The slope, m, is 3.
dy/dx = 3 = 1 + e^(-x)
e^(-x) = 2
-x = ln(2)
x = -ln(2)
y = -ln(2) - 2 = ~ -2.693147
The tangent point is (-ln(2),-ln(2)-2)
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y+ln(2)+2 = 3*(x+ln(2))
y+ln(2)+2 = 3x+3ln(2)
y = 3x + 2ln(2) - 2
y = 3x - 0.6137