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

May 12, 2016

Equation of normal is $\left(y - 8.57664\right) = 0.1133 \left(x + 3\right)$

#### Explanation:

As $f \left(x\right) = {x}^{2} - x \sin x$ at $x = - 3$, $f \left(x\right) = {\left(- 3\right)}^{2} - \left(- 3\right) \sin \left(- 3\right)$

= $9 - \left(- 3\right) \left(- \sin 3\right) = 9 - 3 \times 0.14112 = 8.57664$

Hence we have to find normal at point $\left(- 3 , 8.57664\right)$

Slope of curve is given by $\frac{\mathrm{df}}{\mathrm{dx}} = 2 x - \sin x - x \cos x$

At $x = - 3$ it is $2 \left(- 3\right) - \sin \left(- 3\right) - \left(- 3\right) \left(\cos \left(- 3\right)\right)$

= $- 6 + 0.14112 + 3 \left(- 0.99\right) = - 5.85888 - 2.97 = - 8.82888$

Hence slope of normal is $- \frac{1}{-} 8.82888 = - 0.1133$

Hence equation of normal is

$\left(y - 8.57664\right) = 0.1133 \left(x + 3\right)$