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

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

tangent of

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

y = 5 x - 4

Solution steps

Find the tangent point:

(2, 6)

Find the slope of

f (x) = x^{2} + x : \frac{df ( x )}{d x} = 2 x + 1

Compute the slope of

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

x = 2 : m = 5

Find the line with slope m=

5

and passing through

(2, 6) : y = 5 x - 4

y = 5 x - 4

y^{^{''}} + 3 y^{^{'}} + 3 y = 0

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

\frac{d y}{d x} + 3 x y^{3} = 0

\int_{0}^{r} x arcsin (\frac{x}{2 r}) d x

\int \frac{sec ( x ) + cos ( x )}{8 cos ( x )} d x

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