What is the equation of the line normal to f(x)=√3x3−2x at x=2?

1 Answer
Nallasivam V
Jan 7, 2017

y=−11.9x+4

Explanation:

Given -

y=√3x3−2x

It slope at any point is given by its first derivative.

dydx=9x2−22⋅√3x3−2x

At x=2 The slope is -

dydx=9⋅22−22⋅√3⋅23−2⋅2

dydx=36−22⋅√24−4

dydx=342⋅√20)
dydx=342⋅4.5
dydx=349=3.8
m1=3.8

At x=2 the value of the function is

y=√3⋅23−2⋅2
y=√24−4=√20=4.5

The normal is passing through the point (2,4.5)

The slope of the normal m2=−1m1=−13.8

To find the equation of the normal -

y=mx+c
mx+c=y
(−13.8)(2)+c=4.5
−23.8+c=4.5
−11.9+c=4.5
c=4.5−11.9=8.6−11.9=4
The equation is -
y=−11.9x+4

enter image source here

Related questions
See all questions in Normal Line to a Tangent
Impact of this question
1616 views around the world
You can reuse this answer
Creative Commons License