Mathematics - Application of Derivative Question with Solution | TestHub

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