What is the equation of the line normal to #f(x)= sqrt(2x^3-x) # at #x=1#?

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Answer 1 · 2016-02-19T18:45:01

#y=5/2x-5/2#

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

#y=(2x^3-x)^(1/2)#

Applying the chain rule #rArr#

#y'=1/2(2x^3-x)^(-1/2)xx(6x^2-1)#

#y'=((6x^2-1))/(2sqrt(2x^3-x))#

This equals the gradient #m#.

So when #x=1rArr#

#m=((6-1))/(2sqrt(2-1))=5/2#

The tangent line is of the form:

#y=mx+c#

At #x=1#,

#y=sqrt(2-1)=1#

#:.1=5/2xx1 +c#

#:.c=-5/2#

So the equation of the tangent is:

#y=5/2x-5/2#