Question

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

Answer 1

`y = 1 - x`

Explanation:

Let `y = f(x) = ln(x^2 - x + 1)/x`

`dy/dx = (x-2)/(x^2 (x^2-x+1))+(2-log(x^2-x+1))/x^2`

Find the y coordinate at x = 1:

`y = ln(1^2 - 1 + 1)/1 = 0`

The slope, m, of the tangent line is `dy/dx` evaluated at `x = 1`:

`m = (1-2)/(1^2 (1^2-x+1))+(2-log(1^2-1+1))/1^2`

`m = 1`

The slope, n, of the normal line is:

`n = -1/m`

`n = -1/1`

`n = -1`

The normal line has a slope of -1 and passes through the point (1, 0). Using the point-slope form of the equation of a line, we obtain the following equation:

`y - 0 = -1(x - 1)`

`y = 1 - x`