What is the tangent of f(x)=-(6x)/(x^2+1),\at x=0 ?

The tangent of f(x)=-(6x)/(x^2+1),\at x=0 is y=-6x

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tangent of f(x)=-(6x)/(x^2+1),\at x=0

tangent of

f (x) = - \frac{6 x}{x ^{2} + 1}, at x = 0

y = - 6 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = - \frac{6 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{6 ( - 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 = - 6

Find the line with slope m=

- 6

and passing through

(0, 0) : y = - 6 x

y = - 6 x

What is the tangent of f(x)=-(6x)/(x^2+1),\at x=0 ?
The tangent of f(x)=-(6x)/(x^2+1),\at x=0 is y=-6x