What is the tangent of f(x)=4x^3-4x,(-1,0) ?

The tangent of f(x)=4x^3-4x,(-1,0) is y=8x+8

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^3-4x,(-1,0)

tangent of

f (x) = 4 x^{3} - 4 x, (- 1, 0)

y = 8 x + 8

Solution steps

Find the slope of

f (x) = 4 x^{3} - 4 x : \frac{df ( x )}{d x} = 12 x^{2} - 4

Compute the slope of

f (x) = 4 x^{3} - 4 x

(- 1, 0) : m = 8

Find the line with slope m=

8

and passing through

(- 1, 0) : y = 8 x + 8

y = 8 x + 8

f (x) = 2 e^{2 x^{2} + 3 x}, at x = 0

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

\frac{\partial}{\partial x} (sin (5 π x + 4 y))

\frac{d}{d x} ((\frac{1}{e ^{x}})^{100})

\int \frac{e ^{x}}{( e ^{x} + 1 ) ln ( e ^{x} + 1 )} d x

What is the tangent of f(x)=4x^3-4x,(-1,0) ?
The tangent of f(x)=4x^3-4x,(-1,0) is y=8x+8