What is the equation of the normal line of $f (x) = 6 x^{3} - 12 x^{2} - x - 1$ $f(x)=6x^3-12x^2-x-1$ at $x = 2$ $x=2$ ?

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Answer 1 · 2016-09-07T09:37:21

$x + 23 y + 67 = 0$ $x+23y+67=0$

$f (x) = 6 x^{3} - 12 x^{2} - x - 1$ $f(x)=6x^3-12x^2-x-1$ .

When $x = 2, y = f (2) = 48 - 48 - 2 - 1 = - 3$ $x=2, y=f(2)=48-48-2-1=-3$ .

Thus, normal to $f$ $f$ at #(2,-3) is reqd.

We know that, the slope of the tgt. at the pt. $(2,-3)" is "f'(2)$ .

We have, $f'(x)=18x^2-24x-1 rArr f'(2)=72-48-1=23$ .

Since, normal is $bot$ to the tgt., its slope at $(2,-3)$ is $-1/23$ .

Summing up the information about the normal, it has slope $-1/23,$

&, passes thro. $(2,-3)$ .

Using, Point-Slope Form for the Normal, its. eqn. is given by,

$y-(-3)=-1/23(x-2), or, 23(y+3)+(x-2)=0$ ;

i.e., $x+23y+67=0$

Enjoy Maths.!

What is the equation of the normal line of f(x)=6x^3-12x^2-x-1f(x)=6x3−12x2−x−1f(x)=6x^3-12x^2-x-1 at x=2x=2x=2?