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

The tangent of f(x)=x^3,\at =0 is y=3a_{0}^2x-2a_{0}^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)=x^3,\at =0

tangent of

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

y = 3 a_{0}^{2} x - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3})

Find the slope of

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

Compute the slope of

f (x) = x^{3}

x = a_{0} : m = 3 a_{0}^{2}

Find the line with slope m=

3 a_{0}^{2}

and passing through

(a_{0}, a_{0}^{3}) : y = 3 a_{0}^{2} x - 2 a_{0}^{3}

y = 3 a_{0}^{2} x - 2 a_{0}^{3}

x \to 4 + lim (c x^{2} + 3 x)

\int arccot (6 y) d y

x \to 0 + lim (x^{t a n (x)})

\int ((ln (x))^{2}) d x

4 x^{2} tan (x)

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