#f(x) = 15x^(2/3)+5x# is concave downward for all #x<0#
As Kim suggested a graph should make this apparent (See bottom of this post).
Alternately,
Note that #f(0) = 0#
and checking for critical points by taking the derivative and setting to #0#
we get
#f'(x) = 10x^(-1/3)+5 = 0#
or
#10/x^(1/3) = -5#
which simplifies (if #x <> 0#) to
#x^(1/3) = -2#
#rarr# #x=-8#
At #x=-8#
#f(-8) = 15 (-8)^(2/3) + 5(-8)#
#=15(-2)^2 + (-40)#
#=20#
Since (#-8,20#) is the only critical point (other than (#0,0#) )
and #f(x)# decreases from #x=-8# to #x=0#
it follows that #f(x)# decreases on each side of (#-8,20#), so
#f(x)# is concave downward when #x<0#.
When #x>0# we simply note that
#g(x) = 5x# is a straight line and
#f(x) = 15x^(2/3) +5x# remains a positive amount (namely #15x^(2/3)# above that line
therefore #f(x)# is not concave downward for #x>0#.
graph{15x^(2/3) + 5x [-52, 52, -26, 26]}
