# How do you find the critical numbers of f(x)= 2x^3 + 3x^2-12x?

Nov 14, 2016

So the critical points are $\left(- 1 , 20\right)$ and $\left(1 , - 7\right)$

$\left(- 1 , 20\right)$ is a maximum
$\left(1 , - 7\right)$ is a minium

#### Explanation:

We have $f \left(x\right) = 2 {x}^{3} + 3 {x}^{2} - 12 x$

To identify the critical vales, we differentiate and find find values of $x$ st $f ' \left(x\right) = 0$

 { ( f'(x) < 0, => f(x) " is decreasing" ), ( f'(x) = 0, => f(x) " is stationary" ), ( f'(x) > 0, => f(x) " is increasing" ) :}

Differentiating wrt $x$' we have:

$f ' \left(x\right) = 6 {x}^{2} + 6 x - 12$ .... [1]

At a critical point, $f ' \left(x\right) = 0$

$f ' \left(x\right) = 0 \implies 6 {x}^{2} + 6 x - 12 = 0$
$\therefore {x}^{2} + x - 2 = 0$
$\therefore \left(x + 2\right) \left(x - 1\right) = 0$
$x = - 2 , 1$

Ton find the y-coordinate we substitute the required value into $f \left(x\right)$
$x = - 2 \implies f \left(- 2\right) = 2 \left(- 8\right) + 3 \left(4\right) - 12 \left(- 2\right) = - 16 + 12 + 24 = 20$
$x = 1 \implies f \left(1\right) = 2 + 3 - 12 = - 7$

So the critical points are $\left(- 1 , 20\right)$ and $\left(1 , - 7\right)$

Although this answers the question, let's go a bit further and identify the nature of these critical points by looking at the sign of second derivative, and

 { ( f''(x) < 0, => f'(x) " is decreasing" => "maximum" ), ( f''(x) = 0, => f'(x) " is stationary" => "inflection" ), ( f''(x) > 0, => f'(x) " is increasing" => "minimum" ) :}

Differentiating [1] wrt $x$ gives s;

$f ' ' \left(x\right) = 12 x + 6$
$x = - 2 \implies f ' ' \left(- 2\right) = - 14 + 6 < 0$, ie a maximum
$x = 1 \implies f ' ' \left(1\right) = 12 + 6 > 0$, ie a minimum

Incidental, As this is a cubic with a positive coefficient of ${x}^{3}$, we can deduce that the critical point corresponding to the smallest value of $x$ must be the maximum and that corresponding to the larger value must be the maximum. It is not possible to have any other possibility!