Answer:
[tex](-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x<0 \ or \ x>\frac{16}{3}[/tex]
Step-by-step explanation:
[tex]h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \ respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x[/tex]
For positive rate of change [tex]h'(x)>0[/tex]
[tex]\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x<0 \Rightarrow (\frac{3}{8}x-2)<0 \Rightarrow h'(x)>0 \ (multiplication \ of \ two \ negative \ gives \ positive)\\\\When (\frac{3}{8}x-2)>0 \Rightarrow x>\frac{16}{3} \Rightarrow x>0\\\\h'(x)=x(\frac{3}{8}x-2) >0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x<0 \ or \ x>\frac{16}{3} \\\\[/tex]
