Since $\displaystyle{x}={2}\usepackage{cmbright}\Huge\$ , $\displaystyle\frac{{1}}{{2}}{{x}}^{{4}}-\frac{{3}}{{4}}{{x}}^{{2}}=\frac{{1}}{{2}}\cdot{{2}}^{{4}}-\frac{{3}}{{4}}\cdot{{2}}^{{2}}=\frac{{1}}{{2}}\cdot{16}-\frac{{3}}{{4}}\cdot{4}={8}-{3}={5}\usepackage{cmbright}\Huge\$ .