What is the tangent of f(x)=7+4x^2-2x^3,\at x=1 ?

The tangent of f(x)=7+4x^2-2x^3,\at x=1 is y=2x+7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=7+4x^2-2x^3,\at x=1

tangent of

f (x) = 7 + 4 x^{2} - 2 x^{3}, at x = 1

y = 2 x + 7

Solution steps

Find the tangent point:

(1, 9)

Find the slope of

f (x) = 7 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

f (x) = 7 + 4 x^{2} - 2 x^{3}

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 9) : y = 2 x + 7

y = 2 x + 7

\frac{d}{d x} (c e^{x} + c e^{- x})

\frac{d}{d x} (\frac{x}{x - 1})

\int x^{3} (1 + 9 x^{4}) d x

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

\int 6 x^{13} e^{- x^{7}} d x

What is the tangent of f(x)=7+4x^2-2x^3,\at x=1 ?
The tangent of f(x)=7+4x^2-2x^3,\at x=1 is y=2x+7