What is the tangent of f(x)= 6/(3x+1),(-1,-3) ?

The tangent of f(x)= 6/(3x+1),(-1,-3) is y=-9/2 x-15/2

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tangent of f(x)= 6/(3x+1),(-1,-3)

tangent of

f (x) = \frac{6}{3 x + 1}, (- 1, - 3)

y = - \frac{9}{2} x - \frac{15}{2}

Solution steps

Find the slope of

f (x) = \frac{6}{3 x + 1} : \frac{df ( x )}{d x} = - \frac{18}{( 3 x + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{6}{3 x + 1}

(- 1, - 3) : m = - \frac{9}{2}

Find the line with slope m=

- \frac{9}{2}

and passing through

(- 1, - 3) : y = - \frac{9}{2} x - \frac{15}{2}

y = - \frac{9}{2} x - \frac{15}{2}

\int \frac{1}{2} ln (\frac{1 + x}{1 - x}) d x

\int \frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{4}} d x

n = 1 \sum \infty \frac{1 0 ^{n}}{9 ^{n}}

x \frac{d y}{d x} - y = x^{2} e^{x}

\frac{\partial}{\partial x} (ln (3 x + 5 y))

What is the tangent of f(x)= 6/(3x+1),(-1,-3) ?
The tangent of f(x)= 6/(3x+1),(-1,-3) is y=-9/2 x-15/2