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

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

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

tangent of

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

y = 4

Solution steps

Find the tangent point:

(- 1, 4)

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 = - 1 : m = 0

Find the line with slope m=

0

and passing through

(- 1, 4) : y = 4

y = 4

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

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 10)^{6})

n = 1 \sum \infty \frac{1}{3 ^{n} + 1}

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

\frac{\partial}{\partial y} (- 2 x y - y^{4})

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