Question

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

Answer 1

`y=1+1/4(x-1)=1/4 x+3/4`

Explanation:

The normal line to the graph of `f` at the point `(a,f(a))` is perpendicular ("normal") to the tangent line at that point. The tangent line has slope `f'(a)`, so the normal line has slope `-1/(f'(a))` (the "negative reciprocal" of `f'(a)`).

Hence, the equation of the normal line to `f` at the point `(a,f(a))` has equation

`y=f(a)-1/(f'(a))(x-a)`.

In the present case, with `f(x)=1/(2x^2-1)`, the quotient rule leads to `f'(x)=-(4x)/((2x^{2}-1)^2)`. Hence, `f(1)=1/(2-1)=1` and `f'(1)=-(4)/((2-1)^2)=-4`, making `-1/(f'(a))=1/4`.

This means the answer is `y=1+1/4(x-1)=1/4 x+3/4`.