What is the tangent of f(x)=5+5x^2-2x^3,\at x=1 ?

The tangent of f(x)=5+5x^2-2x^3,\at x=1 is y=4x+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=5+5x^2-2x^3,\at x=1

tangent of

f (x) = 5 + 5 x^{2} - 2 x^{3}, at x = 1

y = 4 x + 4

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

f (x) = 5 + 5 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 10 x

Compute the slope of

f (x) = 5 + 5 x^{2} - 2 x^{3}

x = 1 : m = 4

Find the line with slope m=

4

and passing through

(1, 8) : y = 4 x + 4

y = 4 x + 4

n = 0 \sum \infty \frac{7}{4 ^{n}}

y = (6 x + 5) (x^{2} - 3 x)

\int 8 x cos (4) x^{2} d x

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

What is the tangent of f(x)=5+5x^2-2x^3,\at x=1 ?
The tangent of f(x)=5+5x^2-2x^3,\at x=1 is y=4x+4