What is the tangent of f(x)=(8x)/(x^2+1),(-1,-4) ?

The tangent of f(x)=(8x)/(x^2+1),(-1,-4) is y=-4

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tangent of f(x)=(8x)/(x^2+1),(-1,-4)

tangent of

f (x) = \frac{8 x}{x ^{2} + 1}, (- 1, - 4)

y = - 4

Solution steps

Find the slope of

f (x) = \frac{8 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{8 x}{x ^{2} + 1}

(- 1, - 4) : m = 0

Find the line with slope m=

0

and passing through

(- 1, - 4) : y = - 4

y = - 4

\int_{- 2}^{0} (x + 2)^{2} d x

\frac{x}{x}

2.6 - 3.15 t^{2} + 7.3 t^{3}

x \to 0 lim (sin (6 x))

f (θ) = 3 cos (θ)

What is the tangent of f(x)=(8x)/(x^2+1),(-1,-4) ?
The tangent of f(x)=(8x)/(x^2+1),(-1,-4) is y=-4