Question

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

Answer 1

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]}