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

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Answer 1 · 2016-01-09T04:43:49

$y = - \frac{x}{6} - \frac{11}{3}$ $y=-x/6-11/3$ or $x + 6 y = - 22$ $x+6y=-22$

the given: $f (x) = x^{3} - 6 x$ $f(x) = x^3-6x$ at $x = 2$ $x=2$

Find the value of the function at $x = 2$ $x=2$

$f (2) = {(2)}^{3} - 6 (2) = 8 - 12 = - 4$ $f(2) = (2)^3-6(2)=8-12=-4$

$f (2) = - 4$ $f(2) = -4$

The point on the curve is $(2, - 4)$ $(2, -4)$

Find first derivative $f' (x)$ then evaluate at $x=2$

$f' (x)=3x^2-6$

$f'(2)=3(2)^2-6$

$m=6$ to be used for Tangent line.

$m$ (perpendicular) $=-1/6$ to be used for the Normal Line

Use now the $m=-1/6$ and the point $(2,-4)$

The Normal Line:

$y-y_1=m(x-x_1)$

$y-(-4)=(-1/6)(x-2)$

$y+4=(-1/6)(x-2)$

$6y+24=-x+2$

$6y=-x-22$

$y=-x/6-11/3$ the required Normal Line

What is the equation of the line normal to f(x)=x^3-6x f(x)=x3−6xf(x)=x^3-6x at x=2x=2x=2?