Question

What is the equation of the line normal to `f(x)=1/abs(x)` at `x=1`?

Answer 1

`y=x`

Explanation:

For `x > 0`, we have `absx = x`, so `f(x) = 1/x` and `y=1`.

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

At `x=1` the slope of the tangent line is `f'(1) = -1` the slope of the normal line is the opposite reciprocal, so `m_(norm) = -1/((-1)) = 1`.

The equation of the line through `(1,1)` with slope `1` is

`y=x`.

Answer 2

`y = x`

Explanation:

See image at end of post....

we can see that at `x = 1` we have `1/absx = 1/x`

and so the slope is `d/dx (1/x)\_(x = 1) = (- 1/x^2)_(x = 1) = -1`

Thus the tangent vector is `vec T = ((1), (-1))`

Therefore, for the normal vector `vec N = ((a),(b))` we have

`((-1),(1)) * ((a),(b)) = 0 implies a = b` so we can choose `a = b = 1` so that `vec N = ((1),(1))`

the slope of the normal is therefore `1/1 = 1`

we can also say that `y(1) = 1`

so we have

`m = (y - y_o)/(x - x_o)`

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

or `y = x`