What is the tangent of f(x)=2x^3,\at x=1 ?

The tangent of f(x)=2x^3,\at x=1 is y=6x-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=2x^3,\at x=1

tangent of

f (x) = 2 x^{3}, at x = 1

y = 6 x - 4

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = 2 x^{3} : \frac{df ( x )}{d x} = 6 x^{2}

Compute the slope of

f (x) = 2 x^{3}

x = 1 : m = 6

Find the line with slope m=

6

and passing through

(1, 2) : y = 6 x - 4

y = 6 x - 4

\frac{d}{d t} (\frac{1}{1 + 24 e ^{- 0.28 t}})

x y^{^{'}} + 2 y = 3 x, y (1) = 5

\int \frac{x}{( x ^{2} + 2 ) ^{\frac{4}{3}}} d x

y^{^{'}} = 3 y - y^{2}

c x

What is the tangent of f(x)=2x^3,\at x=1 ?
The tangent of f(x)=2x^3,\at x=1 is y=6x-4