How do you find the relative extrema for #f(x)=(x^4)+(4x^3)-12#?

1 Answer
Dec 30, 2017

#f# has global minimum at #x_0=-3# , #f(-3)=-39#

Explanation:

#f(x)=x^4+4x^3-12#, #D_f=RR#

#f'(x)=4x^3+12x^2=4x^2(x+3)#

#f'(x)=0 <=># #(x=0# , #x=-3)#

#f'(x)>0 <=> 4x^2(x+3)>0# #<=># #x##in##(-3,0)uu(0,+oo)#

and #f# is continuous at #x_0=0# so #f# is strictly increasing in #[-3,+oo)#

#f'(x)<0 <=> 4x^2(x+3)<0# #<=># #x##in##(-oo,-3)#

so #f# is strictly decreasing in #(-oo,-3]#

  • #x<-3# #=># #f(x)>f(-3)=-39# - (#f# decreasing)
  • #x>=-3# #=># #f(x)>=f(-3)=-39# - (#f# increasing)

As a result #f# has relative minimum at #x_0=-3# , #f(-3)=-39# which also is an global minimum.
(you can mention that #f(0)# is a saddle point.)