What is the equation of the line that is normal to f(x)=-3x^2-sinx at x=pi/3?

1 Answer
Dec 28, 2017

y=-pi^2/3-sqrt3/2+2/(4pi+1)*(x-pi/3)

Explanation:

f(x)=−3x^2−sinx
a=π/3


n: y=f(a)-1/(f'(a))*(x-a)


f(pi/3)=-(3(pi/3)^2)-sin(pi/3)=-pi^2/3-sqrt3/2

f'(x)=-6x-cosx

f'(pi/3)=-6pi/3-cos(pi/3)=-2pi-1/2=(-4pi-1)/2=-(4pi+1)/2

n: y=-pi^2/3-sqrt3/2-1/(-(4pi+1)/2)*(x-pi/3)

n: y=-pi^2/3-sqrt3/2+2/(4pi+1)*(x-pi/3)