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

The tangent of f(x)=(1-x)(x^2-3)^2,\at x=2 is y=-9x+17

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=(1-x)(x^2-3)^2,\at x=2

tangent of

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

y = - 9 x + 17

Solution steps

Find the tangent point:

(2, - 1)

Find the slope of

f (x) = (1 - x) (x^{2} - 3)^{2} : \frac{df ( x )}{d x} = - 5 x^{4} + 4 x^{3} + 18 x^{2} - 12 x - 9

Compute the slope of

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

x = 2 : m = - 9

Find the line with slope m=

- 9

and passing through

(2, - 1) : y = - 9 x + 17

y = - 9 x + 17

x \to 2 + lim ((2 - x)^{x - 2})

3 y^{^{''}} - 5 y^{^{'}} - 14 y = 0

\int_{4}^{7} (x - 6)^{2} d x

x \to 5 lim (\frac{x ^{2} - 9}{x - 3})

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

What is the tangent of f(x)=(1-x)(x^2-3)^2,\at x=2 ?
The tangent of f(x)=(1-x)(x^2-3)^2,\at x=2 is y=-9x+17