Question

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

Answer 1

`y=-x+1`

Explanation:

First, find the point the normal line will intercept.

`f(1)=ln(1-1+1)=ln(1)=0`

The normal line will pass through the point `(1,0)`.

To find the slope of the normal line, we must first find the slope of the tangent line. To do this, we need to find the function's derivative.

Use the chain rule:

`d/dx[ln(x)]=1/x" "=>" "d/dx[ln(u)]=1/u*u'=(u')/u`

Here, `u=x^2-x+1` so

`f'(x)=(d/dx[x^2-x+1])/(x^2-x+1)=(2x-1)/(x^2-x+1)`

The slope of the tangent line is equal to the value of the derivative at `x=1`.

`f'(1)=(2-1)/(1-1+1)=1/1=1`

The normal line, however, is perpendicular to the tangent line so its slope will be the opposite reciprocal of the slope of the tangent line, which is `1`. The opposite reciprocal of `1` and slope of the normal line is `-1`.

The normal line's equation can be written in `y=mx+b` form, recalling that `m=-1`:

`y=-x+b`

Substitute the point `(1,0)`:

`0=-1+b" "=>" "b=1`

The equation of the normal line is `y=-x+1`. Graphed are the original function and the normal line:

graph{(y-ln(x^2-x+1))(y+x-1)=0 [-5.555, 8.49, -2.61, 4.41]}