What is the tangent of f(x)=5+5x^2-2x^3,\at x=1 ?

The tangent of f(x)=5+5x^2-2x^3,\at x=1 is y=4x+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=5+5x^2-2x^3,\at x=1

tangent of

f (x) = 5 + 5 x^{2} - 2 x^{3}, at x = 1

y = 4 x + 4

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

f (x) = 5 + 5 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 10 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4

Find the line with slope m=

4

and passing through

(1, 8) : y = 4 x + 4

y = 4 x + 4

\frac{d}{d y} (e^{x - y})

\frac{\partial}{\partial x} (cos (3 x + 6 y))

ma c l a u r in x^{2} e^{x^{2}}

\int e^{6 x} sin (7 x) d x

s l o p e (4, - 5), (6, - 1)

What is the tangent of f(x)=5+5x^2-2x^3,\at x=1 ?
The tangent of f(x)=5+5x^2-2x^3,\at x=1 is y=4x+4