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

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

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

tangent of

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

y = - x - 3

Solution steps

Find the slope of

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

Compute the slope of

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

(- 2, - 1) : m = - 1

Find the line with slope m=

- 1

and passing through

(- 2, - 1) : y = - x - 3

y = - x - 3

\frac{d}{d x} ((6 x^{2} + 7 x)^{4})

e^{x^{2} + x}

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

\int \frac{3}{2 x + 7} d x

\int_{1}^{e} (\frac{1}{x}) d x

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