What is the tangent of f(x)=ln(x^2-3x+1),\at x=3 ?

The tangent of f(x)=ln(x^2-3x+1),\at x=3 is y=3x-9

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tangent of f(x)=ln(x^2-3x+1),\at x=3

tangent of

f (x) = ln (x^{2} - 3 x + 1), at x = 3

y = 3 x - 9

Solution steps

Find the tangent point:

(3, 0)

Find the slope of

f (x) = ln (x^{2} - 3 x + 1) : \frac{df ( x )}{d x} = \frac{2 x - 3}{x ^{2} - 3 x + 1}

Compute the slope of

f (x) = ln (x^{2} - 3 x + 1)

x = 3 : m = 3

Find the line with slope m=

3

and passing through

(3, 0) : y = 3 x - 9

y = 3 x - 9

f (x) = (2 x + 4)^{\frac{1}{5}}

n = 1 \sum \infty 1 2^{n} x^{n} n!

\int_{0}^{\infty} \frac{9}{x ^{2} + 1} d x

x \to 0 lim (\frac{x}{x + sin ( x )})

n = 1 \sum \infty (- \frac{1}{3})^{n}

What is the tangent of f(x)=ln(x^2-3x+1),\at x=3 ?
The tangent of f(x)=ln(x^2-3x+1),\at x=3 is y=3x-9