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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(- 1, - 1)

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}

x = - 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}

\frac{d}{d x} (\frac{e ^{a r c t a n (x)}}{1 + x ^{2}})

\frac{d}{d x} (\frac{csc ( 3 x )}{x ^{3}})

\int \frac{x - 4}{x ^{9}} d x

\frac{d}{d x} (x^{\frac{5}{3}})

2 t e^{- 5 t}

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