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

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

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

tangent of

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

y = 2

Solution steps

Find the tangent point:

(- 1, 2)

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}

x = - 1 : m = 0

Find the line with slope m=

0

and passing through

(- 1, 2) : y = 2

y = 2

t \to 0 lim (\frac{t ^{3} - 3 t}{t})

\frac{d}{d x} ((4 x - 3)^{3})

\frac{\partial}{\partial y} (10 x^{\frac{1}{4}} y^{\frac{3}{4}})

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

n = 0 \sum \infty (1)^{n}

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