What is the tangent of f(x)= 1/(x+6),\at x=3 ?

The tangent of f(x)= 1/(x+6),\at x=3 is y=-1/81 x+4/27

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tangent of f(x)= 1/(x+6),\at x=3

tangent of

f (x) = \frac{1}{x + 6}, at x = 3

y = - \frac{1}{81} x + \frac{4}{27}

Solution steps

Find the tangent point:

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

Find the slope of

f (x) = \frac{1}{x + 6} : \frac{df ( x )}{d x} = - \frac{1}{( x + 6 ) ^{2}}

Compute the slope of

f (x) = \frac{1}{x + 6}

x = 3 : m = - \frac{1}{81}

Find the line with slope m=

- \frac{1}{81}

and passing through

(3, \frac{1}{9}) : y = - \frac{1}{81} x + \frac{4}{27}

y = - \frac{1}{81} x + \frac{4}{27}

x^{3}, 2 x - 15.8 x^{2}

x \to \infty lim ((1 - \frac{1}{x})^{x - 1})

x - 2, 14 - x, x = 2

\int_{0}^{1} (1 - x)^{2} d x

\int 9 cos^{2} (x) tan^{3} (x) d x

What is the tangent of f(x)= 1/(x+6),\at x=3 ?
The tangent of f(x)= 1/(x+6),\at x=3 is y=-1/81 x+4/27