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

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Answer 1 · 2016-05-12T08:50:52

Equation of normal is #(y-8.57664)=0.1133(x+3)#

As #f(x)=x^2-xsinx# at #x=-3#, #f(x)=(-3)^2-(-3)sin(-3)#

= #9-(-3)(-sin3)=9-3xx0.14112=8.57664#

Hence we have to find normal at point #(-3,8.57664)#

Slope of curve is given by #(df)/(dx)=2x-sinx-xcosx#

At #x=-3# it is #2(-3)-sin(-3)-(-3)(cos(-3))#

= #-6+0.14112+3(-0.99)=-5.85888-2.97=-8.82888#

Hence slope of normal is #-1/-8.82888=-0.1133#

Hence equation of normal is

#(y-8.57664)=0.1133(x+3)#