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

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}}

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}{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}

d er i v a t i v e f (x) = \frac{x ^{2}}{x + 5}

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

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

\frac{\partial}{\partial x} (3 x^{2} - 4 y^{3})

\int_{0}^{\frac{π}{2}} cos (3 x) d x

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