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

The tangent of f(x)= 2/(3x+1),\at x=-1 is y=-3/2 x-5/2

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

tangent of

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

y = - \frac{3}{2} x - \frac{5}{2}

Solution steps

Find the tangent point:

(- 1, - 1)

Find the slope of

f (x) = \frac{2}{3 x + 1} : \frac{df ( x )}{d x} = - \frac{6}{( 3 x + 1 ) ^{2}}

Compute the slope of

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

x = - 1 : m = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(- 1, - 1) : y = - \frac{3}{2} x - \frac{5}{2}

y = - \frac{3}{2} x - \frac{5}{2}

x \to \frac{π}{3} lim (7 x cos (x))

\frac{d}{d x} (\frac{7}{4 x + 1})

y = e^{3 - x}

y + (x^{2} + y^{2}) d x - x d y = 0

6 x^{2} + (x^{3} + 2) \frac{d}{d x} y = 0

What is the tangent of f(x)= 2/(3x+1),\at x=-1 ?
The tangent of f(x)= 2/(3x+1),\at x=-1 is y=-3/2 x-5/2