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

Jul 21, 2016

$y - \left(- \tan \left(6 \pi - 2\right)\right) = \frac{x - 3}{2 \pi \cdot {\sec}^{2} \left(6 \pi - 2\right)}$

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 $f \left(x\right)$.

$f ' \left(x\right) = - 2 \pi \cdot {\sec}^{2} \left(2 \pi x - 2\right)$

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.

$y - \left(- \tan \left(6 \pi - 2\right)\right) = \frac{x - 3}{2 \pi \cdot {\sec}^{2} \left(6 \pi - 2\right)}$