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

1 Answer
Jul 30, 2016

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

Explanation:

To identify critical points color(blue)"find g'(x) and equate to zero"

differentiate g(x) using the color(blue)"power rule"

rArrg'(x)=6x^2-24=6(x^2-4)=6(x-2)(x+2)

Equating g'(x) to zero

6(x-2)(x+2)=0rArrx=±2

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

g(-2)=2(-2)^3-24(-2)+5=-16+48+5=37

g(2)=2(2)^3-24(2)+5=16-48+5=-27

rArr(-2,37)" and " (2,-27)" are critical points"

To test for local max/min use the color(red)"second derivative test"

• " If g''(a) > 0 , then local min"

• "If g''(a) < 0 , then local max"

rArrg''(x)=12x

g''(-2)=12(-2)=-24<0rArr(-2,37)" is local max"

g''(2)=12(2)=24>0rArr(2,-27)" is local min"
graph{2x^3-24x+5 [-80, 80, -40, 40]}