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

\int \frac{x}{x ^{2} + 3 x + 3} d x

8 tan^{2} (8 x)

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

\int 3 y^{5} d y

\int_{- a}^{a} f (x) d x

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