What are the global and local extrema of #f(x)=x^3 + 4x^2 - 5x # ?

Question

Aurora Alvarado · Accepted Answer

The function has no global extrema. It has a local maximum of #f((-4-sqrt31)/3) = (308+62sqrt31)/27# and a local minimum of #f((-4+sqrt31)/3) = (308-62sqrt31)/27#

For #f(x)=x^3 + 4x^2 - 5x # ,

#lim_(xrarr-oo)f(x)=-oo# so #f# has no global minimum.

#lim_(xrarroo)f(x)=oo# so #f# has no global maximum.

#f'(x)=3x^2+8x-5# is never undefined and is #0# at

#x=(-4+-sqrt31)/3#

For numbers far from #0# (both positive and negative), #f'(x)# is positive. For numbers in #((-4-sqrt31)/3,(-4+sqrt31)/3)#, 3f'(x)# is negative.

The sign of #f'(x)# changes from + to - as we move past #x=(-4-sqrt31)/3#, so #f((-4-sqrt31)/3)# is a local maximum.

The sign of #f'(x)# changes from - to + as we move past #x=(-4+sqrt31)/3#, so #f((-4+sqrt31)/3)# is a local minimum.

Finish by doing the arithmetic to get the answer:

#f# has a local maximum of #f((-4-sqrt31)/3) = (308+62sqrt31)/27# and a local minimum of #f((-4+sqrt31)/3) = (308-62sqrt31)/27#

Christopher Agresta · Answer

undefined The global maximum occurs at the local maximum, which is at the critical point $ x = -\frac{4}{3} $, and the global minimum occurs at the local minimum, which is at the critical point $ x = 1 $.