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

The tangent of f(x)=x^3,\at x=2 is y=12x-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,\at x=2

tangent of

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

y = 12 x - 16

Solution steps

Find the tangent point:

(2, 8)

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 = 2 : m = 12

Find the line with slope m=

12

and passing through

(2, 8) : y = 12 x - 16

y = 12 x - 16

\int 4 x arctan (x) d x

x \to 9 lim (\frac{1}{x - 9})

\frac{d}{d x} ((x + \frac{2}{x}) (7 - 2 x^{3}))

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

(8 - r^{2}) d r = r^{3} sin (θ) d θ, r (0) = 2

What is the tangent of f(x)=x^3,\at x=2 ?
The tangent of f(x)=x^3,\at x=2 is y=12x-16