# How do you find the maxima and minima of the function f(x)=x^3+3x^2-24x+3?

Jan 30, 2017

maxima at $\left(- 4 , 83\right)$
minima at $\left(2 , - 25\right)$

#### Explanation:

We have:

$f ' \left(x\right) = 3 {x}^{2} + 6 x - 24$

At a max/min (turning point) $f ' \left(x\right) = 0 \implies 3 {x}^{2} + 6 x - 24 = 0$

$\therefore \setminus \setminus \setminus \setminus {x}^{2} + 2 x - 8 = 0$
$\therefore \left(x - 2\right) \left(x + 4\right) = 0$
$\therefore x = - 4 , 2 ,$

When $x = - 4 \implies f \left(- 4\right) = - 64 + 48 + 96 + 3 = 83$
When $x = 2 \setminus \setminus \setminus \setminus \setminus \implies f \left(2\right) \setminus \setminus \setminus \setminus \setminus = 8 + 12 - 48 + 3 \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 25$

To determine the nature of the turning points we look at the second derivative. Differentiating again wrt $x$:

$f ' ' \left(x\right) = 6 x + 6$

When $x = - 4 \implies f ' ' \left(- 4\right) < 0 \implies$ maximum
When $x = 2 \setminus \setminus \setminus \setminus \setminus \implies f ' ' \left(2\right) > 0 \setminus \setminus \setminus \setminus \setminus \setminus \implies$ minimum

So the maxima and minima are:

maxima at $\left(- 4 , 83\right)$
minima at $\left(2 , - 25\right)$

We can confirm these results graphically:
graph{x^3+3x^2-24x+3 [-10, 10, -50, 100.]}