What are the critical points of #f(x) = sqrt((x^2 - 4x +5))#?

Question

2940 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-16T15:38:21

The critical point of#f(x)# is #(2,1)#

#y=sqrt(x^2-4x+5)#

Differentiate

#y'=(2x-4)/(2sqrt(x^2-4x+5))#

now critical points are points at which #y'=0# or the tangent is vertical at that point i.e. #y'=1/0#

#y'=0##color(blue)rarr##2x-4=0#

#x=2#

#y'=1/0##color(blue)rarr##x^2-4x+5=0#

#x=2+-i# which it refused as it's not a real number

So there is one critical point at #x=2#
#f(2)=1#

The critical point of#f(x)# is #(2,1)#