What is the tangent of f(x)= 1/(x+8),\at x=9 ?

The tangent of f(x)= 1/(x+8),\at x=9 is y=-1/289 x+26/289

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tangent of f(x)= 1/(x+8),\at x=9

tangent of

f (x) = \frac{1}{x + 8}, at x = 9

y = - \frac{1}{289} x + \frac{26}{289}

Solution steps

Find the tangent point:

(9, \frac{1}{17})

Find the slope of

f (x) = \frac{1}{x + 8} : \frac{df ( x )}{d x} = - \frac{1}{( x + 8 ) ^{2}}

Compute the slope of

f (x) = \frac{1}{x + 8}

x = 9 : m = - \frac{1}{289}

Find the line with slope m=

- \frac{1}{289}

and passing through

(9, \frac{1}{17}) : y = - \frac{1}{289} x + \frac{26}{289}

y = - \frac{1}{289} x + \frac{26}{289}

n = 2 \sum \infty \frac{1}{n ^{3}}

\int \frac{x}{5 + 4 x - x ^{2}} d x

\int_{1}^{\infty} x^{2} d x

f (x) = \frac{2 x ^{2} + 6 x + 6}{x + 1}

\frac{d y}{d x} = e^{4 y} sin (5 x)

What is the tangent of f(x)= 1/(x+8),\at x=9 ?
The tangent of f(x)= 1/(x+8),\at x=9 is y=-1/289 x+26/289