What is the tangent of f(x)= 6/(5x+2),\at x=-1 ?

The tangent of f(x)= 6/(5x+2),\at x=-1 is y=-10/3 x-16/3

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tangent of f(x)= 6/(5x+2),\at x=-1

tangent of

f (x) = \frac{6}{5 x + 2}, at x = - 1

y = - \frac{10}{3} x - \frac{16}{3}

Solution steps

Find the tangent point:

(- 1, - 2)

Find the slope of

f (x) = \frac{6}{5 x + 2} : \frac{df ( x )}{d x} = - \frac{30}{( 5 x + 2 ) ^{2}}

Compute the slope of

f (x) = \frac{6}{5 x + 2}

x = - 1 : m = - \frac{10}{3}

Find the line with slope m=

- \frac{10}{3}

and passing through

(- 1, - 2) : y = - \frac{10}{3} x - \frac{16}{3}

y = - \frac{10}{3} x - \frac{16}{3}

f (x) = arctan (x)

\int \frac{1}{sin ( θ )} d θ

\int \frac{1}{3 x - 3} d x

\int_{4}^{6} \frac{x}{x ^{2} + 6 x + 13} d x

x^{2} 9 - x^{2}

What is the tangent of f(x)= 6/(5x+2),\at x=-1 ?
The tangent of f(x)= 6/(5x+2),\at x=-1 is y=-10/3 x-16/3