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

The tangent of f(x)=(6x)/(x^2+1),\at x=-1 is y=-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = - 3

Solution steps

Find the tangent point:

(- 1, - 3)

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 = 0

Find the line with slope m=

0

and passing through

(- 1, - 3) : y = - 3

y = - 3

\frac{\partial}{\partial x} (- y sin (x ln (y) ln (y)))

\int \frac{1}{x ^{2} - 6 x - 16} d x

n = 1 \sum \infty e^{- 6 n}

\frac{d}{d x} (2 x + \frac{x ^{2} - x}{x ^{3} + x + 1})

sin (4 x) cos (2 x)

What is the tangent of f(x)=(6x)/(x^2+1),\at x=-1 ?
The tangent of f(x)=(6x)/(x^2+1),\at x=-1 is y=-3