The minimum value of the function y=x⁴-2 x²+1 in the interval (...

($\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}$\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}=$12 x²-4 \therefore\left(\frac{d^2 y}{d x^2}\right)_{x=1}$(=12(1)^2-4=8 \gt 0) (\therefore \quad) there is minimum value$of function at (\mathrm{x}=$1) (\therefore \quad) minimum value is $y(1)=1^4-2(1)^2+1=1-2+1=0$