What is the tangent of f(x)=x^3+2x,\at x=0 ?

The tangent of f(x)=x^3+2x,\at x=0 is y=2x

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+2x,\at x=0

tangent of

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

y = 2 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

f (x) = x^{3} + 2 x

x = 0 : m = 2

Find the line with slope m=

2

and passing through

(0, 0) : y = 2 x

y = 2 x

\frac{\partial}{\partial x} (8 x y + 2 x^{4} + 2 y^{4})

\frac{d}{d x} (sin^{n - 2} (x))

\frac{\partial}{\partial x} (\frac{- cos ( x )}{x})

\frac{0.6 e ^{x}}{cos ( x )}

3^{x}

What is the tangent of f(x)=x^3+2x,\at x=0 ?
The tangent of f(x)=x^3+2x,\at x=0 is y=2x