What is the tangent of f(x)= 1/(x+2),\at x=0 ?

The tangent of f(x)= 1/(x+2),\at x=0 is y=-1/4 x+1/2

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tangent of f(x)= 1/(x+2),\at x=0

tangent of

f (x) = \frac{1}{x + 2}, at x = 0

y = - \frac{1}{4} x + \frac{1}{2}

Solution steps

Find the tangent point:

(0, \frac{1}{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 = 0 : m = - \frac{1}{4}

Find the line with slope m=

- \frac{1}{4}

and passing through

(0, \frac{1}{2}) : y = - \frac{1}{4} x + \frac{1}{2}

y = - \frac{1}{4} x + \frac{1}{2}

\int 9 x^{3} y + 3 y^{2} d x

x \to 0 - lim (4 (\frac{1}{x} - csc (x)))

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

\frac{\partial}{\partial x} (2 y - x^{2} e^{- y})

t^{3} e^{b \cdot t}

What is the tangent of f(x)= 1/(x+2),\at x=0 ?
The tangent of f(x)= 1/(x+2),\at x=0 is y=-1/4 x+1/2