What is the tangent of f(x)=(8x)/(x^2-1),\at x=3 ?

The tangent of f(x)=(8x)/(x^2-1),\at x=3 is y=-5/4 x+27/4

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tangent of f(x)=(8x)/(x^2-1),\at x=3

tangent of

f (x) = \frac{8 x}{x ^{2} - 1}, at x = 3

y = - \frac{5}{4} x + \frac{27}{4}

Solution steps

Find the tangent point:

(3, 3)

Find the slope of

f (x) = \frac{8 x}{x ^{2} - 1} : \frac{df ( x )}{d x} = \frac{8 ( - x ^{2} - 1 )}{( x ^{2} - 1 ) ^{2}}

Compute the slope of

f (x) = \frac{8 x}{x ^{2} - 1}

x = 3 : m = - \frac{5}{4}

Find the line with slope m=

- \frac{5}{4}

and passing through

(3, 3) : y = - \frac{5}{4} x + \frac{27}{4}

y = - \frac{5}{4} x + \frac{27}{4}

(\frac{6 x ^{2}}{8 - x})^{3}

x^{- 5}

\int 4 sin (5 x) d x

\int \frac{1}{x ^{\frac{2}{3}} ( 8 + x ^{\frac{1}{3}} )} d x

y (θ) = ln (θ + 1) - e^{θ}

What is the tangent of f(x)=(8x)/(x^2-1),\at x=3 ?
The tangent of f(x)=(8x)/(x^2-1),\at x=3 is y=-5/4 x+27/4