How do you find all local maximum and minimum points given #y=x^4-2x^2+3#?

1 Answer
Oct 28, 2016

Answer:

#(-1,2) and (-1,2)# is a local minimum
#(0,3)# is a local maximum

Explanation:

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

#y'=0#
#4x(x-1)(x+1)=0#

#4x=0rArrx=0#
#x-1=0rArrx=1#
#x+1=0rArrx=-1#

#color(brown)(y'<0) #
# color(brown)(-oo< x< -1 and 0 < x <1)#

#color(blue)(y'>0)#
#color(blue)(-1< x < 0 and 1 < x <+oo)#

So,
the function
#y# decreases for #color(brown)(-oo < x <-1)# reaching the point #(-1,2)#
#y# increases for #color(blue)(-1< x <0)#
So,#(color(red)(-1,2))# is a local minimum

#y# increases for #color(blue)(-1< x <0)# reaching the point#(0,3)#
then it decreases #color(brown)(0 < x <1) #
So,#(color(red)(0,3))# is a local maximum

#y# decreases for #color(brown)(0 < x <1)# reaching the point#(1,2)#then
#y# increases for #color(blue)(1< x < oo)#
So,#(color(red)(-1,2))# is a local minimum