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}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 0

Find the line with slope m=

0

and passing through

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

y = - 4

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