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

The tangent of f(x)= 1/(3+2x),\at x=2 is y=-2/49 x+11/49

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

tangent of

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

y = - \frac{2}{49} x + \frac{11}{49}

Solution steps

Find the tangent point:

(2, \frac{1}{7})

Find the slope of

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

Compute the slope of

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

x = 2 : m = - \frac{2}{49}

Find the line with slope m=

- \frac{2}{49}

and passing through

(2, \frac{1}{7}) : y = - \frac{2}{49} x + \frac{11}{49}

y = - \frac{2}{49} x + \frac{11}{49}

\int \frac{x ^{2}}{1 + x ^{4}} d x

\frac{d}{d x} (\frac{cos ( sin ( x + 1 ) )}{tan ( \frac{1}{x} )})

\int \frac{1}{1 - 9 x ^{2}} d x

\frac{d y}{d x} = e^{3 x + 2 y}

n \to \infty lim (\frac{n}{6} sin (\frac{6}{n}))

What is the tangent of f(x)= 1/(3+2x),\at x=2 ?
The tangent of f(x)= 1/(3+2x),\at x=2 is y=-2/49 x+11/49