What is the tangent of f(x)=ln(x^2-3x+1),(3,0) ?

The tangent of f(x)=ln(x^2-3x+1),(3,0) is y=3x-9

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tangent of f(x)=ln(x^2-3x+1),(3,0)

tangent of

f (x) = ln (x^{2} - 3 x + 1), (3, 0)

y = 3 x - 9

Solution steps

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)

(3, 0) : m = 3

Find the line with slope m=

3

and passing through

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

y = 3 x - 9

\int_{- 5}^{0} (\frac{y}{5} + y + 6) d y

\frac{d y}{d x}, x y = 84

x \to 2 lim (2 x^{2} + 5 x + 2)

\frac{d}{d x} (e^{- s i n (x)})

\int \frac{( x - 5 ) ^{2}}{x ^{3}} d x

What is the tangent of f(x)=ln(x^2-3x+1),(3,0) ?
The tangent of f(x)=ln(x^2-3x+1),(3,0) is y=3x-9