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

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

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