What is the equation of the line normal to f(x)=−13−2x at x=0?

Explanation:rewrite f(x) as f(x)=(3−2x)−1( using the 'chain rule' ) gives:f′(x)=−1(3−2x)−2⋅ddx(3−2x)⇒f′(x)=−(3−2x)−2(−2)=2(3−2x)2at x=0:f′(0)=232=29and f(0)=1/3For perpendicular lines m1⋅m2=−1⇒m2=−129=−92equation of normal : y - b = m (x - a )where m=−92(a,b)=(0,13)( substitute in values ) y−13=−92x⇒y=−92x+13

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High School
Calculus and Analysis
Calculus 1
Derivatives

Rubi Riggs

Answered question

2022-04-09

What is the equation of the line normal to

f (x) = - \frac{1}{3 - 2 x}

at x=0?

Answer & Explanation

Kapovkah7hpl

Beginner2022-04-10Added 14 answers

Explanation:
rewrite f(x) as

f (x) = {(3 - 2 x)}^{- 1}

( using the 'chain rule' ) gives:

f^{'} (x) = - 1 {(3 - 2 x)}^{- 2} \cdot \frac{d}{d x} (3 - 2 x)

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

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

and f(0)=1/3
For perpendicular lines

m_{1} \cdot 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} (a, b) = (0, \frac{1}{3})

( substitute in values )

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

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

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