#f(x)=2x^(5/3) - 5x^(4/3) = x^(4/3)(2x^(1/3)-5)#,
#f' (x)=10/3x^(1/3)(x^(1/3) - 2)# ,
#f'' (x)=(20(x^(1/3) - 1)) / (9x^(2/3))#
Domain: all real numbers (There is no division and there are no even roots to lead to exceptions.)
#x# intercepts: #x^(4/3)(2x^(1/3)-5)=0# where
#x^(4/3)=0# #" "# or #" "# #2x^(1/3)-5=0#
#x=0# #" "# or #" "# #x^(1/3)=5/2# so #x=125/8#
#y# intercept #f(0) = 0#
There are no vertical asymptotes (#f# is continuous on #RR# so there is no #a# at which #f(x) rarr+-oo#)
Critical Numbers and analysis of #f'#
#f'(x)#is never undefined and is #0# at #0# and at #8#, so the critical numbers for #f# are #0# and #8#.
#f'# is positive on #(-oo,0)# so #f# is increasing.
#f'# is negative on #(0,8)# so #f# is decreasing. And #f(0)=0# is a local max.
#f'# is positive on #(8,oo)# so #f# is increasing. And #f(8) = -16# is a local min.
Analysis of #f''#
#f''(x)# is undefined at #0# and is #0# at #1#
Both #20# and the denominator #9x^(2/3)# are always positive, so the saign of #f''# is the same as that of #x^(1/3)-1#
On #(-oo,0)# and on #(0,1)#, #f'(x)# is negative so #f# is concave down.
On #(1,oo)#, #f'(x)# is positive so #f# is concave up.
The point #(1,-3)# is an inflection point.
Putting all of this analysis together you should get a graph like this. (You can zoom in and out and drag the graph around with a mouse.)
graph{2x^(5/3)-5x^(4/3) [-27.63, 37.27, -17.91, 14.54]}
