# How do you find the exact relative maximum and minimum of the polynomial function of #2x^3 -23x^2+78x-72#?

##### 1 Answer

Oct 11, 2016

There is a minimum at

There is a maximum at

#### Explanation:

Given -

#y=2x^3-23x^2+78x-72#

#dy/dx=6x^2-46x+78#

#(d^2y)/(dx^2)=12x-46#

#dy/dx=0 => 6x^2-46x+78=0#

#x=(-b+-sqrt(b^2 -4ac))/(2xxa)#

#x=(-(-46)+-sqrt((-46^2) -(4xx6xx78)))/(2xx6)#

#x=(46+-sqrt(2116-1872))/(2xx6)#

#x=(46+-sqrt(244))/(12)#

#x=(46+-15.62)/(12)#

#x=(46+15.62)/(12)=61.62/12=5.135#

#x=(46-15.62)/(12)=30.38/12=2.531#

At

#(d^2y)/(dx^2)=12(5.135)-46=61.62-46=15.62 >0#

There is a minimum at

At

#(d^2y)/(dx^2)=12(2.531)-46=30.37-46=-15.628<0#

There is a maximum at