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Q.
The minimum value of the function $y = x ^{4}-2 x ^{2}+1$ in the interval $\left[\frac{1}{2}, 2\right]$ is
Application of Derivatives
Solution:
$\frac{d y}{d x}=\frac{d}{d x}\left(x^{4}-2 x^{2}+1\right)$
$=4 x^{3}-4 x=4 x\left(x^{2}-1\right)$
For max. or min, $\frac{d y}{d x}=0$
$ \Rightarrow 4 x\left(x^{2}-1\right)=0$
either $x =0$ or $x =\pm 1$
$x =0$ and $x =-1$
does not belong to $\left[\frac{1}{2}, 2\right]$
$\frac{ d ^{2} y }{ dx ^{2}}=12 x ^{2}-4$
$ \Rightarrow \left(\frac{ d ^{2} y }{ dx ^{2}}\right)_{ x =1}=12(1)^{2}- 4 = 8 > 0$
$\therefore $ there is minimum value of function at $x=1$
$\therefore $ minimum value is
$y(1)=1^{4}-2(1)^{2}+1=1-2+1=0$