What is the tangent of f(x)=x^4+8x^3-2x-2,\at x=-2 ?

The tangent of f(x)=x^4+8x^3-2x-2,\at x=-2 is y=62x+78

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^4+8x^3-2x-2,\at x=-2

tangent of

f (x) = x^{4} + 8 x^{3} - 2 x - 2, at x = - 2

y = 62 x + 78

Solution steps

Find the tangent point:

(- 2, - 46)

Find the slope of

f (x) = x^{4} + 8 x^{3} - 2 x - 2 : \frac{df ( x )}{d x} = 4 x^{3} + 24 x^{2} - 2

Compute the slope of

f (x) = x^{4} + 8 x^{3} - 2 x - 2

x = - 2 : m = 62

Find the line with slope m=

62

and passing through

(- 2, - 46) : y = 62 x + 78

y = 62 x + 78

\int \frac{sec ^{2} ( x )}{1 + tan ( x )} d x

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

\frac{d}{d x} (\frac{1}{x ^{2} - 9})

\int 3 x e^{4 x} d x

What is the tangent of f(x)=x^4+8x^3-2x-2,\at x=-2 ?
The tangent of f(x)=x^4+8x^3-2x-2,\at x=-2 is y=62x+78