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

x \to 3 lim (\frac{x ^{2} - 2 x + 1}{2})

(3 \frac{y}{x} - 8) d x + 3 d y = 0

y = \frac{1}{sin ^{2} ( x )}

\frac{d y}{d x} = 7 x cos (x^{2})

\frac{d y}{d x} = 8 y, y (6) = 7

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