What is the tangent of f(x)=(6x)/(x^2+1),(-1,-3) ?

The tangent of f(x)=(6x)/(x^2+1),(-1,-3) is y=-3

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tangent of f(x)=(6x)/(x^2+1),(-1,-3)

tangent of

f (x) = \frac{6 x}{x ^{2} + 1}, (- 1, - 3)

y = - 3

Solution steps

Find the slope of

f (x) = \frac{6 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{6 x}{x ^{2} + 1}

(- 1, - 3) : m = 0

Find the line with slope m=

0

and passing through

(- 1, - 3) : y = - 3

y = - 3

\int r \cdot e^{- r^{2}} d r

\int \frac{16 - x ^{2}}{x} d x

\int_{- 1}^{0} 8 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

y^{^{'}} = - \frac{x y}{x ^{2} + y ^{2}}

\int \frac{x ^{2}}{1 + x ^{2}} d x

What is the tangent of f(x)=(6x)/(x^2+1),(-1,-3) ?
The tangent of f(x)=(6x)/(x^2+1),(-1,-3) is y=-3