How do you find the exact relative maximum and minimum of the polynomial function of #f(x) = x^3 + 4x^2 - 5x#?

1 Answer
Dec 25, 2017

minimum: #~~(0.523,-1.378)#

maximum: #~~(-3.189,24.193)#

Explanation:

relative max/min can only exist if #f'(x)# is 0 at that x-value.

first find #f'(x)#:
#=3x^2+8x-5# (power rule)

solving this quadratic: #x=(-8+-sqrt(8^2-4(3)(-5)))/(2*3)#
#x=(-8+-sqrt(124))/6#
#x=(-8+-2sqrt(31))/6#
#x=(-4+-sqrt(31))/3#

since #f(x)# is a cubic with a positive leading coefficient, it will approach #-oo# on the left and #+oo# on the right. this means the lesser x-value where #f'(x)=0# , (#(-4-sqrt(31))/3#) , is a local maximum and the greater x-value , (#(-4+sqrt(31))/3#) , is a local minimum.

plugging the x-values into #f(x)# gives:
minimum: #(-4/3+sqrt(31)/3,308/27-(62sqrt(31))/27)# #~~(0.523,-1.378)#

maximum: #(-4/3-sqrt(31)/3,308/27+(62sqrt(31))/27)# #~~(-3.189,24.193)#