# What are the absolute extrema of  f(x)= 6x^3 − 9x^2 − 36x + 3 in [-4,8]?

Jun 9, 2018

$\left(- 4 , - 381\right)$ and $\left(8 , 2211\right)$

#### Explanation:

In order to find the extrema, you need to take the derivative of the function and find the roots of the derivative.

i.e. solve for $\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right] = 0$ , use power rule:

$\frac{d}{\mathrm{dx}} \left[6 {x}^{3} - 9 {x}^{2} - 36 x + 3\right] = 18 {x}^{2} - 18 x - 36$

solve for the roots:
$18 {x}^{2} - 18 x - 36 = 0$
${x}^{2} - x - 2 = 0$ , factor the quadratic:
$\left(x - 1\right) \left(x + 2\right) = 0$
$x = 1 , x = - 2$

$f \left(- 1\right) = - 6 - 9 + 36 + 3 = 24$
$f \left(2\right) = 48 - 36 - 72 + 3 = - 57$

Check the bounds:
$f \left(- 4\right) = - 381$
$f \left(8\right) = 2211$

Thus the absolute extrema are $\left(- 4 , - 381\right)$ and $\left(8 , 2211\right)$