Question

What is the slope of the line normal to the tangent line of `f(x) = sin(2x-pi/6)` at `x= pi/3`?

Answer 1

No slope

Explanation:

Find the derivative:

`f(x)=sin(2x-pi/6)`
`f'(x)=cos(2x-pi/6)x`

Substitute `x=pi/3` to find the slope
`f'(pi/3)=cos(2(pi/3)-pi/6)(pi/3)=cos(pi/2)(pi/3)=0`

The slope of a normal line is negative reciprocal of the derivative.
`-1/0` is undefined then the it must be a vertical line.