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

Question

1566 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-28T16:03:50

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

#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)#