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

Dec 28, 2017

$y = - {\pi}^{2} / 3 - \frac{\sqrt{3}}{2} + \frac{2}{4 \pi + 1} \cdot \left(x - \frac{\pi}{3}\right)$

#### Explanation:

f(x)=−3x^2−sinx
a=π/3

$n : y = f \left(a\right) - \frac{1}{f ' \left(a\right)} \cdot \left(x - a\right)$

$f \left(\frac{\pi}{3}\right) = - \left(3 {\left(\frac{\pi}{3}\right)}^{2}\right) - \sin \left(\frac{\pi}{3}\right) = - {\pi}^{2} / 3 - \frac{\sqrt{3}}{2}$

$f ' \left(x\right) = - 6 x - \cos x$

$f ' \left(\frac{\pi}{3}\right) = - 6 \frac{\pi}{3} - \cos \left(\frac{\pi}{3}\right) = - 2 \pi - \frac{1}{2} = \frac{- 4 \pi - 1}{2} = - \frac{4 \pi + 1}{2}$

$n : y = - {\pi}^{2} / 3 - \frac{\sqrt{3}}{2} - \frac{1}{- \frac{4 \pi + 1}{2}} \cdot \left(x - \frac{\pi}{3}\right)$

$n : y = - {\pi}^{2} / 3 - \frac{\sqrt{3}}{2} + \frac{2}{4 \pi + 1} \cdot \left(x - \frac{\pi}{3}\right)$