What is the tangent of f(x)=x^3-4x^2,\at x=0 ?

The tangent of f(x)=x^3-4x^2,\at x=0 is y=0

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^3-4x^2,\at x=0

tangent of

f (x) = x^{3} - 4 x^{2}, at x = 0

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = x^{3} - 4 x^{2} : \frac{df ( x )}{d x} = 3 x^{2} - 8 x

Compute the slope of

f (x) = x^{3} - 4 x^{2}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

\frac{d}{d x} (lo g_{e} (3))

n = 2 \sum \infty \frac{1}{n ( n + 3 )}

\frac{d y}{d x} = \frac{2 x}{y + x ^{2} y}

ln (\frac{3 + e ^{x}}{3 - e ^{x}})

\int (4 - x^{2})^{2} x^{2} d x

What is the tangent of f(x)=x^3-4x^2,\at x=0 ?
The tangent of f(x)=x^3-4x^2,\at x=0 is y=0