Question

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

Answer 1

Let's start by simplifying the function.

`f(x) = 1/sqrt((x - 1)(x - 1))`

`f(x) = 1/(x- 1)`

Now, let's start by finding the corresponding y-coordinate to `x = 2`.

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

Next, let's differentiate the function.

`f(x) = 1/(x - 1)`

By the quotient rule:

`f'(x) = (0 xx (x - 1) - 1(1))/(x- 1)^2`

`f'(x) = -1/(x -1)^2`

The slope of the tangent (the line that touches the graph at the point `x = a`) at the point `x= 2` is

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

The normal line is perpendicular to the tangent. So, the slope of the normal line is `-1/(-1) = 1`

We now know the normal line's point of contact on the function `(2, 1)` and the slope of the normal line. We can finally use point-slope form to determine the equation of the normal line.

`y - y_1 = m(x- x_1)`

`y - 1 = 1(x- 2)`

`y - 1 = x - 2`

`y = x - 1`

Hence, the equation of the normal line is `y = x- 1`.

Hopefully this helps!