Find the derivative of each critical point and determine the local extreme values?

y= x(4 - x^2)

The answer should be:

When critical point: x= 1.15; derivative=0; extremum = local max; and value=3.08

Critical point: x= -1.15; derivative= 0; extremum = local min; value= -3.08 ;

But Idk how to get this! Can someone help me??? What are the correct steps?

Question

Find the derivative of each critical point and determine the local extreme values? 

y= x(4 - x^2) 

The answer should be: 

When critical point: x= 1.15; derivative=0;  extremum = local max; and value=3.08 

Critical point: x= -1.15; derivative= 0; extremum = local min; value= -3.08 ; 

But Idk how to get this! Can someone help me??? What are the correct steps?

anonymous · Answer

$$y= x(4 - x^2) =4x-x^3$$ is a start
then 
$$y'=4-3x^2$$

anonymous · Answer

find the critical points by solving 
$$4-3x^2=0$$