What is the tangent of f(x)=(1-x)(x^2-8)^2,\at x=3 ?

The tangent of f(x)=(1-x)(x^2-8)^2,\at x=3 is y=-25x+73

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^2-8)^2,\at x=3

tangent of

f (x) = (1 - x) (x^{2} - 8)^{2}, at x = 3

y = - 25 x + 73

Solution steps

Find the tangent point:

(3, - 2)

Find the slope of

f (x) = (1 - x) (x^{2} - 8)^{2} : \frac{df ( x )}{d x} = - 5 x^{4} + 4 x^{3} + 48 x^{2} - 32 x - 64

Compute the slope of

f (x) = (1 - x) (x^{2} - 8)^{2}

x = 3 : m = - 25

Find the line with slope m=

- 25

and passing through

(3, - 2) : y = - 25 x + 73

y = - 25 x + 73

f (x) = 7 arcsin (x^{4})

\int \frac{6 x + 7}{x ^{2} + 4 x + 4} d x

\int_{- 3}^{1} x^{2} e^{x^{3} + 27} d x

\frac{( 3 s + 3 )}{s ^{2} - 9}

\int cos^{3} (6 x) d x

What is the tangent of f(x)=(1-x)(x^2-8)^2,\at x=3 ?
The tangent of f(x)=(1-x)(x^2-8)^2,\at x=3 is y=-25x+73