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 \frac{tan ( x )}{csc ( x ) + cot ( x )} d x

\frac{d}{d x} (4 x^{5} + 2 x^{3} + 6 x^{2} + 3 x)

\int 8 e^{2 θ} sin (2 θ) d θ

y^{^{'}} + y = x \cdot e^{- x} + 1

2 x^{3} + 36 x^{2} + 210 x + 11

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