# What is the equation of the line normal to #f(x)= sqrt(3x^3-2x) # at #x=2#?

##### 1 Answer

Jan 7, 2017

#y=-1/1.9x+4#

#### Explanation:

Given -

#y=sqrt(3x^3-2x)#

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

#dy/dx=(9x^2-2)/(2* sqrt(3x^3-2x)#

At

#dy/dx=(9*2^2-2)/(2 * sqrt(3*2^3-2*2))#

#dy/dx=(36-2)/(2*sqrt(24-4))#

#dy/dx=34/(2*sqrt20))#

#dy/dx=34/(2*4.5)#

#dy/dx=34/9=3.8#

#m_1=3.8#

At

#y=sqrt(3*2^3-2*2)#

#y=sqrt(24-4)=sqrt 20=4.5#

The normal is passing through the point

The slope of the normal

To find the equation of the normal -

#y=mx+c#

#mx+c=y#

#(-1/3.8)(2)+c=4.5#

#-2/3.8+c=4.5#

#-1/1.9+c=4.5#

#c=4.5-1/1.9=(8.6-1)/1.9=4#

The equation is -

#y=-1/1.9x+4#