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

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

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

tangent of

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

y = - 2

Solution steps

Find the slope of

f (x) = \frac{- 4 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

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

(1, - 2) : m = 0

Find the line with slope m=

0

and passing through

(1, - 2) : y = - 2

y = - 2

\int \frac{x}{e ^{- 3 x}} d x

\int tan^{2} (w) d w

\frac{\partial}{\partial x} (- a e^{k z} w sin (k x - wt))

y^{^{''}} - 5 y^{^{'}} = 25 t, y (0) = 5, y^{^{'}} (0) = 0

\int_{0}^{2} x 4 - x^{2} d x

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