How do you find the critical points #f(x)= x^3 + 35x^2 - 125x - 9375#?

1 Answer
Jun 27, 2015

Answer:

The critical numbers are: #x = -25# and #x=5/3#

Explanation:

A critical number for #f# is a number in the domain of #f# at which, #f' =0# or #f'# does not exist.

#f(x)= x^3 + 35x^2 - 125x - 9375#

#f'(x) = 3x^2+70x-125#

Clearly, this function exists for all #x#, so we need only consider its zeros:

#3x^2+70x-125 = 0#

To look for factors using whole numbers:
#3 xx -125= - 375#
Find two numbers whose product is #-375# and whose sum is #70#

#-1 xx 375# won't work
#-2# is not a factor of 375
#-3 xx 125# won't work
#-5 xx 75# STOP! that's the one.
Now split the middle term and factor by grouping:

#3x^2+70x-125 = 0#

#3x^2-5x+75x-125 = 0#

#x(3x-5)+25(3x-5) = 0#

#(x+25)(3x-5) = 0#

#x = -25# and #x=5/3#. Both are in the domain of #f# (Domain of #f = RR#), so they are the critical numbers for #f#.