Find the equation of a tangent line at a give point?
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Find the equation of a tangent line at a give point?

[From: ] [author: ] [Date: 14-03-22] [Hit: ]
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ln(x^2 - 3x + 1) = y at the given point (3,0)

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Derivative of ln(x^2 - 3x + 1) = 1/(x^2 - 3x + 1) * (2x - 3)

Plug in 3 to find the slope of the tangent line at (3, 0):

1/(3^2 - 3(3) + 1) * (2*3 - 3) = 3

y - 0 = 3(x - 3)

y = 3x - 9 <---answer

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f ` (x) = (2x - 3) / (x^2 - 3x + 1)

f ` (3) = 3

y = 3 ( x - 3 )

y = 3 x - 9
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