How do you find the intervals of increasing and decreasing using the first derivative given #y=x^4-2x^2#?

Question

1166 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-30T14:44:07

#y(x)# decreases in the intervals #(-oo,-1)# and #(0,1)# and increases in the intervals #(-1,0)# and #(1,+oo)#.

#y(x)# increases in intervals where #y'(x)>0# and decreases in intervals where #y'(x)<0#

#y'(x) =4x^3-4x=4x(x^2-1)#

Analyze the sign of the derivative considering the two factors separately:

#4x>0# for # x>0#

#(x^2-1) > 0# for # -oo < x < -1 # and # 1 < x < +oo#

So:

#-oo < x < -1 " " y'(x) <0#
#-1 < x < 0 " " y'(x) >0#
#0 < x < 1 " " y'(x) <0#
#1 < x < +oo " " y'(x) >0#

graph{x^4-2x^2 [-10, 10, -5, 5]}