What is the equation of the line normal to #f(x)= -1/(3-2x) # at #x=0#?

1 Answer
Jan 17, 2016

# y = -9/2 x + 1/3 #

Explanation:

rewrite f(x) as f(x) = # (3 - 2x )^-1 #

( using the 'chain rule' ) gives:

# f'(x) = - 1 (3 - 2x )^-2 . d/dx ( 3 - 2x ) #

#rArr f'(x) = - (3 - 2x )^-2 (-2) = 2/(3 - 2x )^2 #

at x = 0 :# f'(0) = 2/3^2 = 2/9#

and f(0) = 1/3

For perpendicular lines # m_1.m_2 = - 1 rArr m_2 = (-1)/(2/9) = -9/2 #

equation of normal : y - b = m (x - a )

where # m = -9/2 ( a , b ) = ( 0 , 1/3 )#

( substitute in values ) : # y - 1/3 = -9/2 x #

#rArr y = -9/2 x + 1/3 #