\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^4-2 \mathrm{x}^2+1\right)=4 \mathrm{x}\left(\mathrm{x}^2-1\right)\)
For \(\max\). or \(\min , \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
\(4 \mathrm{x}\left(\mathrm{x}^2-1\right)=0 ;\), either \(\mathrm{x}=0\) or \(\mathrm{x}= \pm 1\)
\(x=0\) and \(x=-1\) does not belong to \(\left[\frac{1}{2}, 2\right]\)
\(\frac{d^2 y}{d x^2}=12 x^2-4 \therefore\left(\frac{d^2 y}{d x^2}\right)_{x=1}\)
\(=12(1)^2-4=8 > 0\)
\(\therefore \quad\) there is minimum value of function at \(\mathrm{x}=1\)
\(\therefore \quad\) minimum value is
\(y(1)=1^4-2(1)^2+1=1-2+1=0\)