What is the tangent of f(x)=2x^3-x^2+6,\at x=2 ?

The tangent of f(x)=2x^3-x^2+6,\at x=2 is y=20x-22

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=2x^3-x^2+6,\at x=2

tangent of

f (x) = 2 x^{3} - x^{2} + 6, at x = 2

y = 20 x - 22

Solution steps

Find the tangent point:

(2, 18)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 20

Find the line with slope m=

20

and passing through

(2, 18) : y = 20 x - 22

y = 20 x - 22

\int \frac{e ^{x}}{4 - e ^{2 x}} d x

g (x) = 7 + 4 x

\frac{d}{d x} (\frac{5 x}{x ^{2} + 6})

\frac{3 V}{π \cdot r ^{2}}

x \to 0 lim (ln (e^{x}))

What is the tangent of f(x)=2x^3-x^2+6,\at x=2 ?
The tangent of f(x)=2x^3-x^2+6,\at x=2 is y=20x-22