What is the tangent of f(x)=x^3-9x,\at x=(-3) ?

The tangent of f(x)=x^3-9x,\at x=(-3) is y=18x+54

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^3-9x,\at x=(-3)

tangent of

f (x) = x^{3} - 9 x, at x = (- 3)

y = 18 x + 54

Solution steps

Find the tangent point:

((- 3), 0)

Find the slope of

f (x) = x^{3} - 9 x : \frac{df ( x )}{d x} = 3 x^{2} - 9

Compute the slope of

f (x) = x^{3} - 9 x

x = (- 3) : m = 18

Find the line with slope m=

18

and passing through

((- 3), 0) : y = 18 x + 54

y = 18 x + 54

f (x) = 2 x - \frac{1}{x}, at x = 0.5

x \to 1 + lim (\frac{5}{x - 1})

\int \frac{e ^{5 x}}{( e ^{5 x} + 11 ) ^{2}} d x

y^{^{'''}} - y^{^{''}} + 2 y^{^{'}} + 4 y = 0

\int_{- 1}^{1} e^{x} cos (x) d x

What is the tangent of f(x)=x^3-9x,\at x=(-3) ?
The tangent of f(x)=x^3-9x,\at x=(-3) is y=18x+54