What is the tangent of f(x)=7x^2-x^3,(1,6) ?

The tangent of f(x)=7x^2-x^3,(1,6) is y=11x-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=7x^2-x^3,(1,6)

tangent of

f (x) = 7 x^{2} - x^{3}, (1, 6)

y = 11 x - 5

Solution steps

Find the slope of

f (x) = 7 x^{2} - x^{3} : \frac{df ( x )}{d x} = 14 x - 3 x^{2}

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 = 11

Find the line with slope m=

11

and passing through

(1, 6) : y = 11 x - 5

y = 11 x - 5

What is the tangent of f(x)=7x^2-x^3,(1,6) ?
The tangent of f(x)=7x^2-x^3,(1,6) is y=11x-5