Question$$h(z)=\frac{1}{z}+9 z^{2} \text { for } z>0$$
Select the exact global maximum and minimum values of the function. The global maximum of $h(z)$ on $z>0$ does not exist, the global minimum is $\sqrt[3]{18}+\sqrt[3]{\frac{9}{4}}$ The global maximum of $h(z)$ on $z>0$ is $\frac{1}{9}+729$, the global minimum is $\sqrt[3]{9}+\sqrt[3]{\frac{9}{4}}$ The global maximum of $h(z)$ on $z>0$ is $\frac{1}{18}+729$, the global minimum is $\sqrt[3]{18}+\sqrt[3]{\frac{9}{4}}$ The global maximum of $h(z)$ on $z>0$ does not exist, the global minimum is $\sqrt[3]{18}+\sqrt[3]{\frac{9}{2}}$ The global maximum of $h(z)$ on $z>0$ is $\frac{1}{18}+729$, the global minimum is $\sqrt[3]{9}+\sqrt[3]{\frac{9}{4}}$