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

To find the equation of the line normal to the function f(x) at x=1, we need to find the derivative of f(x) and evaluate it at x=1. Then, we can determine the slope of the normal line by taking the negative reciprocal of the derivative at that point. Finally, we can use the point-slope form of a line to write the equation of the line.

First, let's find the derivative of f(x). Using the quotient rule, we have:

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

Next, we evaluate f'(x) at x=1:

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

Simplifying this expression will give us the slope of the normal line at x=1.

Once we have the slope, we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point of tangency (in this case, x=1, f(1)), and m is the slope of the normal line.

By substituting the values into the point-slope form, we can obtain the equation of the line normal to f(x) at x=1.