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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

f (x) = \frac{- 4 x}{x ^{2} + 1}, at x = - 1

y = 2

Solution steps

Find the tangent point:

(- 1, 2)

Find the slope of

f (x) = \frac{- 4 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{- 4 x}{x ^{2} + 1}

x = - 1 : m = 0

Find the line with slope m=

0

and passing through

(- 1, 2) : y = 2

y = 2

\int_{0}^{1} \frac{2}{x ^{2} + 1} d x

\frac{d}{d x} (x^{4} + y^{4} + z^{4})

\frac{d}{d x} (sin (7 x^{2}))

x \frac{d y}{d x} = 7 x e^{x} - y + 9 x^{2}

x \to 1 lim (\frac{2}{x ^{2} - 7 x + 6})

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