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

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

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=1

tangent of

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

y = - 3 x + 2

Solution steps

Find the tangent point:

(1, - 1)

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 = 1 : m = - 3

Find the line with slope m=

- 3

and passing through

(1, - 1) : y = - 3 x + 2

y = - 3 x + 2

\int 9 x^{4} (x^{5} + 2)^{8} d x

- 2 e^{- x^{2}}

y = \frac{csc ( 4 x )}{ln ( x )}

x^{2} \frac{d ^{2} y}{d x ^{2}} + 4 x \frac{d y}{d x} + 2 y = 8 x

y = x^{2} - 8

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