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}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

\frac{d}{d x} (2 x + \frac{200}{x})

d er i v a t i v e f (x) = \frac{32}{0}

\frac{\partial}{\partial y} (e^{- y} cos (π x))

\int 4 tan^{2} (x) sec^{2} (x) d x

\int arcsin (\frac{1}{x}) d x

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