we have the function
[tex]f(x)=x^3-\frac{3}{2}x^2[/tex]Find out the first derivative of the given function
[tex]f^{\prime}(x)=3x^2-3x[/tex]Equate to zero the first derivative
[tex]\begin{gathered} 3x^2-3x=0 \\ 3x(x-1)=0 \end{gathered}[/tex]The values of x are
x=0 and x=1
we have the intervals
(-infinite,0) (0,1) (1,infinite)
Interval (-infinite,0) -----> f'(x) is positive
interval (0,1) ---------> f'(x) is negative
interval (1,infinite) -----> f'(x) is positive
that means
x=0 is a local maximum
x=1 is a local minimum
Find out the y-coordinates of the extreme values
For x=0 -----> substitute in the function f(x) ---------> f(x)=0
For x=1 ------> substitute in the function f(x) ------> f(x)=-0.5
therefore
The extreme values are
