What is the tangent of f(x)= 1/(x+2) ?

The tangent of f(x)= 1/(x+2) is y=-1/((a_{0)+2)^2}x+(2a_{0}+2)/((a_{0)+2)^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)= 1/(x+2)

tangent of

f (x) = \frac{1}{x + 2}

y = - \frac{1}{( a _{0} + 2 ) ^{2}} x + \frac{2 a _{0} + 2}{( a _{0} + 2 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0} + 2})

Find the slope of

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

Compute the slope of

f (x) = \frac{1}{x + 2}

x = a_{0} : m = - \frac{1}{( a _{0} + 2 ) ^{2}}

Find the line with slope m=

- \frac{1}{( a _{0} + 2 ) ^{2}}

and passing through

(a_{0}, \frac{1}{a _{0} + 2}) : y = - \frac{1}{( a _{0} + 2 ) ^{2}} x + \frac{2 a _{0} + 2}{( a _{0} + 2 ) ^{2}}

y = - \frac{1}{( a _{0} + 2 ) ^{2}} x + \frac{2 a _{0} + 2}{( a _{0} + 2 ) ^{2}}

x \to 0 lim (2 x (1 - x))

\int_{1}^{\infty} x^{- \frac{1}{2}} d x

\int \frac{( u - 1 )}{( u + 1 )} d u

n = 1 \sum \infty 6 \frac{1}{n}

f (x) = x^{2} + 3 x + 4 π, at x = - 3

What is the tangent of f(x)= 1/(x+2) ?
The tangent of f(x)= 1/(x+2) is y=-1/((a_{0)+2)^2}x+(2a_{0}+2)/((a_{0)+2)^2}