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]}
