What is the equation of the normal line of #f(x)=x^3/(3x^2 + 7x - 1 # at #x=-1 #?

To find the equation of the normal line of f(x) at x=-1, we need to find the slope of the tangent line at x=-1 and then determine the negative reciprocal of that slope to find the slope of the normal line.

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

f'(x) = [(3x^2 + 7x - 1)(3x^2) - (x^3)(6x + 7)] / (3x^2 + 7x - 1)^2

Next, we substitute x=-1 into f'(x) to find the slope of the tangent line at x=-1:

f'(-1) = [(3(-1)^2 + 7(-1) - 1)(3(-1)^2) - ((-1)^3)(6(-1) + 7)] / (3(-1)^2 + 7(-1) - 1)^2

Simplifying this expression, we get:

f'(-1) = -8/81

Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, the slope of the normal line is:

m = -1 / (-8/81) = 81/8

Now, we have the slope of the normal line and the point (-1, f(-1)) = (-1, -1/9). Using the point-slope form of a line, we can write the equation of the normal line:

y - y1 = m(x - x1)

Substituting the values, we get:

y - (-1/9) = (81/8)(x - (-1))

Simplifying this equation, we have:

y + 1/9 = (81/8)(x + 1)

Multiplying through by 8 to eliminate fractions, we get:

8y + 8/9 = 81(x + 1)

Rearranging the terms, we have the equation of the normal line:

81x - 8y + 73 = 0