What is the tangent of f(x)= 1/(x+2),\at x=3 ?

The tangent of f(x)= 1/(x+2),\at x=3 is y=-1/25 x+8/25

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tangent of f(x)= 1/(x+2),\at x=3

tangent of

f (x) = \frac{1}{x + 2}, at x = 3

y = - \frac{1}{25} x + \frac{8}{25}

Solution steps

Find the tangent point:

(3, \frac{1}{5})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{25}

Find the line with slope m=

- \frac{1}{25}

and passing through

(3, \frac{1}{5}) : y = - \frac{1}{25} x + \frac{8}{25}

y = - \frac{1}{25} x + \frac{8}{25}

\frac{\partial}{\partial x} (\frac{1}{1 - x ^{4}})

(3 + x) d x = (e^{t} - e^{- t}) d t

\int (x^{1.1} + 9 x^{3.5}) d x

\int 2 x^{2} cos (x) d x

\frac{d}{d x} (4 arctan (2 x))

What is the tangent of f(x)= 1/(x+2),\at x=3 ?
The tangent of f(x)= 1/(x+2),\at x=3 is y=-1/25 x+8/25