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}}

Compute the slope of

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

x = 3 : 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} (e^{x} cos (x))

f (x) = x^{2} + 5

\frac{d}{d x} (cos (2 x + π))

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

(9 - 2 x^{3})^{3}

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