What is the tangent of f(x)=x^3-4x+6,\at x=2 ?

The tangent of f(x)=x^3-4x+6,\at x=2 is y=8x-10

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^3-4x+6,\at x=2

tangent of

f (x) = x^{3} - 4 x + 6, at x = 2

y = 8 x - 10

Solution steps

Find the tangent point:

(2, 6)

Find the slope of

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

Compute the slope of

f (x) = x^{3} - 4 x + 6

x = 2 : m = 8

Find the line with slope m=

8

and passing through

(2, 6) : y = 8 x - 10

y = 8 x - 10

\int \frac{x ^{2} + 1}{( x - 3 ) ( x - 2 ) ^{2}} d x

f (x) = (- x^{2} + 9)^{2}

\int cosh (3 x) d x

x \frac{d y}{d x} - y = x^{2} - y^{2}

\int sin^{3} (x) cos^{\frac{3}{2}} (x) d x

What is the tangent of f(x)=x^3-4x+6,\at x=2 ?
The tangent of f(x)=x^3-4x+6,\at x=2 is y=8x-10