# How do you find the critical points and local max and min for g(x)=2x^3-24x+5?

Jul 30, 2016

local max at (-2 ,37)
local min at (2 ,-27)

#### Explanation:

To identify critical points $\textcolor{b l u e}{\text{find g'(x) and equate to zero}}$

differentiate g(x) using the $\textcolor{b l u e}{\text{power rule}}$

$\Rightarrow g ' \left(x\right) = 6 {x}^{2} - 24 = 6 \left({x}^{2} - 4\right) = 6 \left(x - 2\right) \left(x + 2\right)$

Equating g'(x) to zero

6(x-2)(x+2)=0rArrx=±2

To find critical points, substitute x = ± 2 into g(x)

$g \left(- 2\right) = 2 {\left(- 2\right)}^{3} - 24 \left(- 2\right) + 5 = - 16 + 48 + 5 = 37$

$g \left(2\right) = 2 {\left(2\right)}^{3} - 24 \left(2\right) + 5 = 16 - 48 + 5 = - 27$

$\Rightarrow \left(- 2 , 37\right) \text{ and " (2,-27)" are critical points}$

To test for local max/min use the $\textcolor{red}{\text{second derivative test}}$

• " If g''(a) > 0 , then local min"

• "If g''(a) < 0 , then local max"

$\Rightarrow g ' ' \left(x\right) = 12 x$

$g ' ' \left(- 2\right) = 12 \left(- 2\right) = - 24 < 0 \Rightarrow \left(- 2 , 37\right) \text{ is local max}$

$g ' ' \left(2\right) = 12 \left(2\right) = 24 > 0 \Rightarrow \left(2 , - 27\right) \text{ is local min}$
graph{2x^3-24x+5 [-80, 80, -40, 40]}