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

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

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tangent of f(x)=(-8x)/(x^2+1),(1,-4)

tangent of

f (x) = \frac{- 8 x}{x ^{2} + 1}, (1, - 4)

y = - 4

Solution steps

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}

(1, - 4) : m = 0

Find the line with slope m=

0

and passing through

(1, - 4) : y = - 4

y = - 4

y^{^{''}} + 3.26 y^{^{'}} + 2.66 y = 0

\frac{\partial}{\partial x} (e^{x} (x + 2 y))

\frac{\partial}{\partial y} (x^{2} - 6 y)

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (1) = 0, y^{^{'}} (1) = 1

\int \frac{3 + u}{u ^{2} - 1} d u

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