Note that f(0)=0
and checking for critical points by taking the derivative and setting to 0
we get
f'(x)=10x-â…“+5=0
or 10x⅓=−5
which simplifies (if x<>0) to
x⅓=−2
→ x=−8
At x=−8
f(−8)=15(−8)23+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)=15xâ…”+5x remains a positive amount (namely 15xâ…”above that line
therefore f(x) is not concave downward for x>0.
Â
Â
Â
 .
Â
Â
Â
Â
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)=15xâ…”+5xÂ
 remains a positive amount (namely  15x⅔ above that line therefore Â
f(x)is not concave downward for  x>0
