What is the tangent of f(x)=x^3-3x,(2,2) ?

The tangent of f(x)=x^3-3x,(2,2) is y=9x-16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^3-3x,(2,2)

tangent of

f (x) = x^{3} - 3 x, (2, 2)

y = 9 x - 16

Solution steps

Find the slope of

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

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 = 9

Find the line with slope m=

9

and passing through

(2, 2) : y = 9 x - 16

y = 9 x - 16

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

\int 2 + \frac{1}{x ^{2}} d x

\frac{d}{d x} ((1 - 4 x)^{- 1})

g (x) = 6^{x}

\frac{\partial}{\partial x} (7 x^{3} y^{2})

What is the tangent of f(x)=x^3-3x,(2,2) ?
The tangent of f(x)=x^3-3x,(2,2) is y=9x-16