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

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

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

tangent of

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

y = - 4

Solution steps

Find the tangent point:

(- 1, - 4)

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}

x = - 1 : m = 0

Find the line with slope m=

0

and passing through

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

y = - 4

\frac{d}{d x} (I n (x^{3}))

\frac{d}{d x} ((x^{2} + x^{3}) \cdot (3 x^{4} - x^{3}))

y^{^{''}} - 2 y^{^{'}} + y = t + 1

x \to 0 lim (\frac{1}{0})

\frac{d}{d x} (- 5 cos (cos (x^{6})))

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