What is the tangent of f(x)= 1/(x+2),\at x=7 ?

The tangent of f(x)= 1/(x+2),\at x=7 is y=-1/81 x+16/81

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tangent of f(x)= 1/(x+2),\at x=7

tangent of

f (x) = \frac{1}{x + 2}, at x = 7

y = - \frac{1}{81} x + \frac{16}{81}

Solution steps

Find the tangent point:

(7, \frac{1}{9})

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 = 7 : m = - \frac{1}{81}

Find the line with slope m=

- \frac{1}{81}

and passing through

(7, \frac{1}{9}) : y = - \frac{1}{81} x + \frac{16}{81}

y = - \frac{1}{81} x + \frac{16}{81}

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

\frac{d ^{4} y}{d x ^{4}} + 2 \frac{d ^{2} y}{d x ^{2}} + y = 0

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

\int_{4}^{9} \frac{1}{9 - x} d x

\frac{\partial}{\partial x} (3 x + 8 y + 2)

What is the tangent of f(x)= 1/(x+2),\at x=7 ?
The tangent of f(x)= 1/(x+2),\at x=7 is y=-1/81 x+16/81