What is the tangent of f(x)= 1/(x+3),\at x=2 ?

The tangent of f(x)= 1/(x+3),\at x=2 is y=-1/25 x+7/25

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tangent of f(x)= 1/(x+3),\at x=2

tangent of

f (x) = \frac{1}{x + 3}, at x = 2

y = - \frac{1}{25} x + \frac{7}{25}

Solution steps

Find the tangent point:

(2, \frac{1}{5})

Find the slope of

f (x) = \frac{1}{x + 3} : \frac{df ( x )}{d x} = - \frac{1}{( x + 3 ) ^{2}}

Compute the slope of

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

x = 2 : m = - \frac{1}{25}

Find the line with slope m=

- \frac{1}{25}

and passing through

(2, \frac{1}{5}) : y = - \frac{1}{25} x + \frac{7}{25}

y = - \frac{1}{25} x + \frac{7}{25}

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What is the tangent of f(x)= 1/(x+3),\at x=2 ?
The tangent of f(x)= 1/(x+3),\at x=2 is y=-1/25 x+7/25