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

Question

Ezekiel Alexander · Accepted Answer

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

Cameron Alexander · Answer

undefined To find the critical points of $ f(x) = x^3 + 35x^2 - 125x - 9375 $, first, take the derivative of the function to find its critical points. Then, set the derivative equal to zero and solve for $ x $. Finally, evaluate the second derivative at each critical point to determine whether they are maximum, minimum, or inflection points.