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

The tangent of f(x)=(x^2-2)(x^3-3x),\at x=-2 is y=26x+48

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^2-2)(x^3-3x),\at x=-2

tangent of

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

y = 26 x + 48

Solution steps

Find the tangent point:

(- 2, - 4)

Find the slope of

f (x) = (x^{2} - 2) (x^{3} - 3 x) : \frac{df ( x )}{d x} = 5 x^{4} - 15 x^{2} + 6

Compute the slope of

f (x) = (x^{2} - 2) (x^{3} - 3 x)

x = - 2 : m = 26

Find the line with slope m=

26

and passing through

(- 2, - 4) : y = 26 x + 48

y = 26 x + 48

x \to 0 lim (\frac{x ^{2} - 5}{x - 2})

\frac{d ^{2}}{d x ^{2}} x y + 2 x^{2} = 5

\int sin^{3} (x) \cdot cos (x) d x

\frac{7}{x + 4}

What is the tangent of f(x)=(x^2-2)(x^3-3x),\at x=-2 ?
The tangent of f(x)=(x^2-2)(x^3-3x),\at x=-2 is y=26x+48