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

Jan 17, 2016

$y = - \frac{9}{2} x + \frac{1}{3}$

Explanation:

rewrite f(x) as f(x) = ${\left(3 - 2 x\right)}^{-} 1$

( using the 'chain rule' ) gives:

$f ' \left(x\right) = - 1 {\left(3 - 2 x\right)}^{-} 2 . \frac{d}{\mathrm{dx}} \left(3 - 2 x\right)$

$\Rightarrow f ' \left(x\right) = - {\left(3 - 2 x\right)}^{-} 2 \left(- 2\right) = \frac{2}{3 - 2 x} ^ 2$

at x = 0 :$f ' \left(0\right) = \frac{2}{3} ^ 2 = \frac{2}{9}$

and f(0) = 1/3

For perpendicular lines ${m}_{1.} {m}_{2} = - 1 \Rightarrow {m}_{2} = \frac{- 1}{\frac{2}{9}} = - \frac{9}{2}$

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

where $m = - \frac{9}{2} \left(a , b\right) = \left(0 , \frac{1}{3}\right)$

( substitute in values ) : $y - \frac{1}{3} = - \frac{9}{2} x$

$\Rightarrow y = - \frac{9}{2} x + \frac{1}{3}$