#f(x) = (1+x)^(2/3)*(3-x)^(1/3)#

By inspection, #f(x)# has zeros at #x=-1# and #x=3#

Applying the product rule:

#f'(x)# = #(1+x)^(2/3) * 1/3(3-x)^(-2/3) * (-1)+#

#2/3(1+x)^(-1/3) * (3-x)^(1/3)#

For points of extrema #f'(x)=0#

I.e. where: #(1+x)^(2/3) * 1/3(3-x)^(-2/3) * (-1)+#

#2/3(1+x)^(-1/3) * (3-x)^(1/3) =0#

#1/3(1=x)^(2/3) * (3-x)^(-2/3) = 2/3(1+x)^(-1/3) * (3-x)^(1/3)#

#(1+x)^(2/3)/((3-x)^(2/3)) = (2*(3-x)^(1/3))/(1+x)^(1/3)#

Cross multiply:

#(1+x)^(2/3+1/3) = 2* (3-x)^(1/3+2/3)#

#1+x = 2(3-x) -> 1+x = 6-2x#

#3x=5#

#x=5/3#

Hence #f(x)# has an extreme value at #x=5/3#

Now consider the graph of #f(x)# below:

graph{(1+x)^(2/3)* (3-x)^(1/3) [-10, 10, -5, 5]}

We can see that #f(x)# has a local maximum at #x=5/3#

#:. f_max(x) = f(5/3) ~= 2.117#

Finally, considering the zeros of #f(x)# we can see that #f(x)# has a discontinuity at #(-1,0)#