What is the tangent of f(x)=6+5x^2-2x^3,\at x=2 ?

The tangent of f(x)=6+5x^2-2x^3,\at x=2 is y=-4x+18

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=6+5x^2-2x^3,\at x=2

tangent of

f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = 2

y = - 4 x + 18

Solution steps

Find the tangent point:

(2, 10)

Find the slope of

f (x) = 6 + 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

(2, 10) : y = - 4 x + 18

y = - 4 x + 18

\int x (3 + 5 x) d x

x \to 2 lim (3 f (x) + 2 g (x) (x)^{2})

\frac{2 s ^{2} + 5 s + 5}{( s + 1 ) ^{3}}

(\frac{5}{x}), at x = 4

\int 3 e^{x} + 5 sin (x) d x

What is the tangent of f(x)=6+5x^2-2x^3,\at x=2 ?
The tangent of f(x)=6+5x^2-2x^3,\at x=2 is y=-4x+18