What is the equation of the line normal to f(x)=x^3-6x f(x)=x36x at x=2x=2?

1 Answer

y=-x/6-11/3y=x6113 or x+6y=-22x+6y=22

Explanation:

the given: f(x) = x^3-6xf(x)=x36x at x=2x=2

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

f(2) = (2)^3-6(2)=8-12=-4f(2)=(2)36(2)=812=4

f(2) = -4f(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