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

1 Answer
Jul 21, 2016

#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))#