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

The tangent of f(x)=(x^2-3x+2)/(x^2-1) is y= 3/((a_{0)+1)^2}x+(a_{0}^2-4a_{0}-2)/((a_{0)+1)^2}

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

tangent of

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

y = \frac{3}{( a _{0} + 1 ) ^{2}} x + \frac{a _{0}^{2} - 4 a _{0} - 2}{( a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0} - 2}{a _{0} + 1})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = \frac{3}{( a _{0} + 1 ) ^{2}}

Find the line with slope m=

\frac{3}{( a _{0} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0} - 2}{a _{0} + 1}) : y = \frac{3}{( a _{0} + 1 ) ^{2}} x + \frac{a _{0}^{2} - 4 a _{0} - 2}{( a _{0} + 1 ) ^{2}}

y = \frac{3}{( a _{0} + 1 ) ^{2}} x + \frac{a _{0}^{2} - 4 a _{0} - 2}{( a _{0} + 1 ) ^{2}}

y = 2.8414 x^{2} - 1.2187 x + 0.5399

\frac{d}{d x} ((x^{2} + x + 2)^{\frac{2}{3}})

y = x^{3} + 1

arcsin (\frac{x}{4})

x \to \infty lim (x^{2 x})

What is the tangent of f(x)=(x^2-3x+2)/(x^2-1) ?
The tangent of f(x)=(x^2-3x+2)/(x^2-1) is y= 3/((a_{0)+1)^2}x+(a_{0}^2-4a_{0}-2)/((a_{0)+1)^2}