Question

What is the equation of the line normal to `f(x)=- tan(2pix -2)` at `x=3`?

Answer 1

`y-(-tan(6pi-2)) = (x-3)/(2pi*sec^2(6pi-2))`

Explanation:

`f(x)=−tan(2πx−2)`
`f(3) =−tan(6π−2)`

Point: `(3, −tan(6π−2))`

To find the slope, we need to take the derivative of `f(x)`.

`f'(x) = -2pi*sec^2(2pix-2)`

The slope normal/perpendicular would be `1/(2pi*sec^2(2pix-2)`.

Therefore at x=3, the perpendicular slope would be `1/(2pi*sec^2(6pi-2)`.

Now, we just need to plugin everything in the point-slope form.

`y-(-tan(6pi-2)) = (x-3)/(2pi*sec^2(6pi-2))`