What is the equation of the tangent line of #f(x)=x^3-12x/(x^2+9x)-1# at #x=-3#?

To find the equation of the tangent line of the function f(x) = (x^3 - 12x) / (x^2 + 9x) - 1 at x = -3, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the quotient rule:

f'(x) = [(x^2 + 9x)(3x^2 - 12) - (x^3 - 12x)(2x + 9)] / (x^2 + 9x)^2

Next, we substitute x = -3 into f'(x) to find the slope of the tangent line:

f'(-3) = [((-3)^2 + 9(-3))(3(-3)^2 - 12) - ((-3)^3 - 12(-3))(2(-3) + 9)] / ((-3)^2 + 9(-3))^2

Simplifying the expression, we get:

f'(-3) = -6/9

Therefore, the slope of the tangent line at x = -3 is -6/9, which simplifies to -2/3.

To find a point on the tangent line, we substitute x = -3 into the original function f(x):

f(-3) = (-3^3 - 12(-3)) / (-3^2 + 9(-3)) - 1

Simplifying the expression, we get:

f(-3) = -27/6 - 1

f(-3) = -29/6

Therefore, a point on the tangent line is (-3, -29/6).

Using the point-slope form of a line, we can write the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values we found, we get:

y - (-29/6) = (-2/3)(x - (-3))

Simplifying the equation, we have:

y + 29/6 = (-2/3)(x + 3)

Multiplying through by 6 to eliminate the fractions, we get:

6y + 29 = -4(x + 3)

Finally, simplifying the equation, we have:

6y + 29 = -4x - 12

Therefore, the equation of the tangent line of f(x) = (x^3 - 12x) / (x^2 + 9x) - 1 at x = -3 is 4x + 6y + 41 = 0.