What is the tangent of f(x)=-2x^3,(2,-16) ?

The tangent of f(x)=-2x^3,(2,-16) is y=-24x+32

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,(2,-16)

tangent of

f (x) = - 2 x^{3}, (2, - 16)

y = - 24 x + 32

Solution steps

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}

(2, - 16) : m = - 24

Find the line with slope m=

- 24

and passing through

(2, - 16) : y = - 24 x + 32

y = - 24 x + 32

\frac{d y}{d x} = (x + y + 7)^{2}

\int_{- r}^{r} r^{2} - x^{2} d x

\frac{d}{d x} (\frac{8}{1 + 8 x})

x \to - 1 lim (2 x^{2} + 3)

x \to - 3 lim (∣ x + 1 ∣ + \frac{3}{x})

What is the tangent of f(x)=-2x^3,(2,-16) ?
The tangent of f(x)=-2x^3,(2,-16) is y=-24x+32