What are the critical values, if any, of #f(x)=(x^2-5x)/(x^2-2)#?

1 Answer
Nov 4, 2015

Answer:

#f# has no critical values.

Explanation:

#f(x)=(x^2-5x)/(x^2-2)#

Note that the domain of #f# is all real numbers except #+-sqrt2#

#f'(x) = ((2x-5)(x^2-2)-(x^2-5x)(2x))/(x^2-2)^2#

# = (2x^3-5x^2-4x+10-2x^3+10x^2)/(x^2-2)^2#

# = (5x^2-4x+10)/(x^2-2)^2#

#f'(x)# is not defined for #x=+-sqrt2#, but those numbers are not in the domain of #f#, so they are not critical values for #f#.

#f'(x)=0# when #5x^2-4x+10 =0#. The discriminant of this quadratic is negative, so there are no real zeros.

#f# has no critical numbers.