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:

$y=5/2x-5/2$

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