How do you find the critical numbers of #f(x)=x^2-6x#?

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Answer 1 · 2015-12-21T07:23:06

Critical point exist at #x=3#

#f(x) = x^2 - 6x#

Critical points can be found at x = c where #f'(c) = 0# critical points are also obtained at #x= c# if #f(c)# is defined and #f'(c)# is not defined.

So first we find derivative #f'(x) #

#f'(x) = 2x - 6#

We find x by solving #f'(x)=0#

#2x - 6 = 0#
#=> 2x = 6 #
#=> (2x)/2 = 6/2#

#=> x= 3# Critical point exists at this value of x.