`y = ln(x^2 - 3x + 1), (3,0)` Find an equation of the tangent line to the curve at the given point.

Textbook Question

Chapter 3, 3.6 - Problem 33 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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The equation of the tangent line to the curve `y = ln (x^2 - 3x + 1)` , at the point (3,0) is the following, such that:

`f(x) - f(x_0) = f'(x_0)(x - x_0)`

You need to put ` ` `f(x) = y, f(x_0) = 0, x_0 = 3`

You need to evaluate f'(3), hence, you need to find the derivative of the function, using the product rule, such that:

`f'(x) = (ln (x^2 - 3x + 1))'*(x^2 - 3x + 1)'`

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

Replacing 3 for x, yields:

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

`f'(3) = 3/1 => f'(3) = 3`

Replacing the values in the equation of tangent line, yields:

`y - 0 = 3*(x - 3)`

`y = 3*(x - 3)`

Hence, evaluating the equation of the tangent line to the given curve, at the point (3,0), yields `y = 3*(x - 3).`

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