How do you find the critical points for #f(x)=x^(8)# and the local max and min?

1 Answer
Jan 5, 2018

Answer:

Check below.

Explanation:

#f(x)=x^8#

#D_f=RR#

#f# is continuous in #RR#

#f'(x)=8x^7# and

#f'(x)=0 <=> x=0#
#x_0=0# is a critical point.
There is no other critical point because #f'# is defined in #RR#.

If #x<0# then #f'(x)<0# so #f# is decreasing in #(-oo,0]#
If #x>0# then #f'(x)>0# so #f# is increasing in #[0,+oo)#

  • #0##<##x# #<=> f(x)>f(0)=0#

  • #x>=0# #<=> f(x)>=f(0)=0#

So for #x##in##RR#: #f(x)>=f(0)#

Therefore #f# has global minimum at #x_0=0# , #f(0)=0#