Question

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

Answer 1

`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`