What is the equation of the line normal to $f(x)=6x^3+2x$ at $x=3$ ?

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Answer 1 · 2016-12-18T03:52:10

$y=-x/164+27555/164$

Given -

$y=6x^3+2x$

$dy/dx=18x^2+2$

Slope, exactly at $x=3$

At $x=3; 18(3)^2+2=164$

This is the slope of the tangent $m_1=164$ .
Slope of the normal $m_2=-1/(m_1)=-1/164$

At $x=3; 6(3)^3+2(3)=162+6=168$

We have to find the equation of the line passing through the point $(3, 168)$ with the slope $-1/164$

$mx+c=y$
$-1/164(3)+c=168$
$c=168+3/164=(27552+3)/164=27555/164$

$y=-x/164+27555/164$

What is the equation of the line normal to f(x)=6x^3+2x f(x)=6x^3+2x at x=3x=3?