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

1 Answer
Mar 25, 2016

Answer:

#(0,0)# is a local minimum and #(4/3,32/27)# is a local maximum.
There are no global extrema.

Explanation:

First multiply the brackets out to make differentiating easier and get the function in the form
#y=f(x)=2x^2-x^3#.

Now local or relative extrema or turning points occur when the derivative #f'(x)=0#,
that is, when #4x-3x^2=0#,
#=> x(4-3x)=0#
#=>x=0 or x=4/3#.

#therefore f(0)=0(2-0)=0 and f(4/3)=16/9(2-4/3)=32/27#.

Since the second derivative #f''(x)=4-6x# has the values of
#f''(0)=4>0 and f''(4/3)=-4<0#,
it implies that #(0,0)# is a local minimum and #(4/3,32/27)# is a local maximum.

The global or absolute minimum is #-oo# and the global maximum is #oo#, since the function is unbounded.

The graph of the function verifies all these calculations :

graph{x^2(2-x) [-7.9, 7.9, -3.95, 3.95]}