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

The tangent of f(x)= 1/(2+3x),\at x=2 is y=-3/64 x+7/32

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

tangent of

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

y = - \frac{3}{64} x + \frac{7}{32}

Solution steps

Find the tangent point:

(2, \frac{1}{8})

Find the slope of

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

Compute the slope of

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

x = 2 : m = - \frac{3}{64}

Find the line with slope m=

- \frac{3}{64}

and passing through

(2, \frac{1}{8}) : y = - \frac{3}{64} x + \frac{7}{32}

y = - \frac{3}{64} x + \frac{7}{32}

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What is the tangent of f(x)= 1/(2+3x),\at x=2 ?
The tangent of f(x)= 1/(2+3x),\at x=2 is y=-3/64 x+7/32