What is the tangent of f(x)= 3/(4x+1),(-1,-1) ?

The tangent of f(x)= 3/(4x+1),(-1,-1) is y=-4/3 x-7/3

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tangent of f(x)= 3/(4x+1),(-1,-1)

tangent of

f (x) = \frac{3}{4 x + 1}, (- 1, - 1)

y = - \frac{4}{3} x - \frac{7}{3}

Solution steps

Find the slope of

f (x) = \frac{3}{4 x + 1} : \frac{df ( x )}{d x} = - \frac{12}{( 4 x + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{3}{4 x + 1}

(- 1, - 1) : m = - \frac{4}{3}

Find the line with slope m=

- \frac{4}{3}

and passing through

(- 1, - 1) : y = - \frac{4}{3} x - \frac{7}{3}

y = - \frac{4}{3} x - \frac{7}{3}

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

x + y^{2} = 20, x + y = 0

f (x) = x^{2} - 3 x + 4

x \to - 8 lim (4 x - 2)

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

What is the tangent of f(x)= 3/(4x+1),(-1,-1) ?
The tangent of f(x)= 3/(4x+1),(-1,-1) is y=-4/3 x-7/3