What are the critical points of #f(x) =(x^2-4x+5)/(x-3)#?

1 Answer
Feb 4, 2016

Answer:

The critical numbers are #1# and #7#.

Explanation:

#f'(x) = ((2x-4)(x-3)-(x^2-4x+5)(1))/(x-3)^2#

# = (x^2-6x+7)/(x-3)^2#

#f'# fails to exists only at #x=3# which is not in the domain of #f#,

and #f'(x) = 0# at #x=1# and at #x=7#, both of which are in the domain of #f#.

So, the critical numbers for #f# are #1# and #7#.

There appears to be some variability in the use of the term "critical point".
I use it to mean a point in the domain of a function at which the derivative is #0# or fails to exist.
Some people seem to use it to mean a point on the graph of a function at which the derivative is #0# or fails to exist.