What is the tangent of f(x)= 6/(4x+1),\at x=-1 ?

The tangent of f(x)= 6/(4x+1),\at x=-1 is y=-8/3 x-14/3

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tangent of f(x)= 6/(4x+1),\at x=-1

tangent of

f (x) = \frac{6}{4 x + 1}, at x = - 1

y = - \frac{8}{3} x - \frac{14}{3}

Solution steps

Find the tangent point:

(- 1, - 2)

Find the slope of

f (x) = \frac{6}{4 x + 1} : \frac{df ( x )}{d x} = - \frac{24}{( 4 x + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{6}{4 x + 1}

x = - 1 : m = - \frac{8}{3}

Find the line with slope m=

- \frac{8}{3}

and passing through

(- 1, - 2) : y = - \frac{8}{3} x - \frac{14}{3}

y = - \frac{8}{3} x - \frac{14}{3}

3 x^{5} - 3 x^{4} + x^{3}

y = \frac{( x ^{3} )}{3 ^{x}}

5 x - x^{2}, y = 0

t^{2} - 2 t + 2

t^{22} ln ∣ t ∣

What is the tangent of f(x)= 6/(4x+1),\at x=-1 ?
The tangent of f(x)= 6/(4x+1),\at x=-1 is y=-8/3 x-14/3