Question

What is the slope of the line normal to the tangent line of `f(x) = 2x-4/sqrt(x-1)` at `x= 2`?

Answer 1

The normal has equation `x = 2`.

Explanation:

If we plug the value of `x` into the function, we get:

`f(2) = 2(2) - 4/sqrt(2 - 1) = 4 - 4/1 = 0`

We now find the derivative.

`f'(x) = 2 - 4/(2sqrt(x - 1))`

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

So at `x = 2`, the tangent would have slope.

`f'(2) = 2 - 2/sqrt(2 - 1) = 2 - 2 = 0`

This would be a horizontal line. Since the normal line is perpendicular to the tangent, the normal line will be vertical, in the form `x = a`. Our initial `x` value was `x = 2`, which will be the equation of the normal.

Hopefully this helps!