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

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

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

tangent of

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

y = - 1

Solution steps

Find the tangent point:

(- 1, - 1)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = 0

Find the line with slope m=

0

and passing through

(- 1, - 1) : y = - 1

y = - 1

\frac{2 e ^{- s}}{( s + \frac{1}{5} )}

y^{^{''}} - 5 y^{^{'}} - 14 y = 0

\frac{\partial}{\partial x} (6)

\frac{\partial}{\partial y} (3 x^{2} e^{y})

\frac{d}{d x} (9 x - 2)

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