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]}
