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}

\frac{\partial}{\partial x} (\frac{3}{3 x + y})

\int - \frac{1}{2} y d y

x \to - 1 lim (x + 3)

\frac{0.6}{s + 2}

\frac{d y}{d x} = \frac{( x ^{2} y - y )}{( y + 1 )}, y (0) = 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