Determine the local max and/or min and intervals of increase and decrease for the function f(x)=√(x^2 - 2x +2) ?

1 Answer
Jim S
May 7, 2018

#f# is decreasing in #(-oo,1]# and increasing in #[1,+oo)# so #f# has a local and global #min# at #x_0=1# , #f(1)=1#
#-> f(x)>=f(1)=1>0# , #x##in##RR#

Explanation:

#f(x)=sqrt(x^2-2x+2)# , #D_f=RR#

#AA##x##in##RR#,

#f'(x)=((x^2-2x+2)')/(2sqrt(x^2-2x+2)# #=#

#(2x-2)/(2sqrt(x^2-2x+2)# #=#

#(x-1)/(sqrt(x^2-2x+2)#

with #f'(x)=0 <=> (x=1)#

  • #x##in##(-oo,1)#, #f'(x)<0# so #f# is decreasing in #(-oo,1]#
  • #x##in##(1,+oo)# , #f'(x)>0# so #f# is increasing in #[1,+oo)#

#f# is decreasing in #(-oo,1]# and increasing in #[1,+oo)# so #f# has a local and global #min# at #x_0=1# , #f(1)=1#
#-> f(x)>=f(1)=1>0# , #x##in##RR#

Graphical help
graph{sqrt(x^2-2x+2) [-10, 10, -5, 5]}

Related questions
Impact of this question
952 views around the world
You can reuse this answer
Creative Commons License