What is the tangent of f(x)= 1/(x+2),\at x=9 ?

The tangent of f(x)= 1/(x+2),\at x=9 is y=-1/121 x+20/121

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tangent of f(x)= 1/(x+2),\at x=9

tangent of

f (x) = \frac{1}{x + 2}, at x = 9

y = - \frac{1}{121} x + \frac{20}{121}

Solution steps

Find the tangent point:

(9, \frac{1}{11})

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 = 9 : m = - \frac{1}{121}

Find the line with slope m=

- \frac{1}{121}

and passing through

(9, \frac{1}{11}) : y = - \frac{1}{121} x + \frac{20}{121}

y = - \frac{1}{121} x + \frac{20}{121}

x^{s i n (x)}

\int cos (x) cos (3 x) d x

\frac{d}{d x} (\frac{x ^{2}}{x + 1})

\frac{6 t}{6 + t}

\frac{d y}{d x} = x y + 9 x + 2 y + 18

What is the tangent of f(x)= 1/(x+2),\at x=9 ?
The tangent of f(x)= 1/(x+2),\at x=9 is y=-1/121 x+20/121