Question

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

Answer 1

x=1

Explanation:

By definition, the normal line must have a slope that is the opposite reciprocal of the slope of the tangent line at `x=1`.
Thus, to determine the slope of the normal line, we must first calculate the slope of the tangent line at x=1, which is just the derivative of `f` at `x=1` or `f'(1)`.

Step 1: Calculate `dy/dx`
`d/dx[(x-1)^2/(x^2+2)]`
`= (2(x^2+2)(x-1)-(x-1)^2(2x))/(x^2+2)^2`

Step 2: Find f'(1)
`f'(1)=(2(3)(0)-(0)^2(2))/(1^2+2)^3`
`f'(1)=0`

Step 3: Determine the equation of the normal line
Knowing that `f'(1)=0` tells us that there is the graph of `f` has a horizontal tangent line at `x=1`. Thus, the normal line must have a slope of `oo`, which is undefined. Because vertical lines are the only type of line with undefined slopes, `f` must have a vertical tangent line at `x=1`.

The equation for the vertical tangent line is `x=1`.