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

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

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

tangent of

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

y = - \frac{2}{25} x + \frac{7}{25}

Solution steps

Find the tangent point:

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

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

Find the line with slope m=

- \frac{2}{25}

and passing through

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

y = - \frac{2}{25} x + \frac{7}{25}

\int (4 x + 2) d x

ln (x x^{2} + 7)

\int \frac{( cos ( x ) )}{sin ( x )} d x

\frac{1}{t}

\frac{d}{d x} (\frac{x + 1}{x - 1})

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