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The minimum value of $2 x^2+x-1$ is :
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The correct answer is:
$-\frac{9}{8}$
Let $y=2 x^2+x-1$
$y^{\prime}=4 x+1$
For maxima or minima, put $y^{\prime}=0$
$\Rightarrow \quad x=-\frac{1}{4}$
$y^{\prime \prime}=4=+\mathrm{ve}$
$y$ is minimum at $x=-\frac{1}{4}$.
Thus, minimum value $=2\left(-\frac{1}{4}\right)^2+\left(-\frac{1}{4}\right)-1$
$\frac{2}{16}-\frac{1}{4}-1=\frac{1}{8}-\frac{5}{4}=-\frac{9}{8}$
$y^{\prime}=4 x+1$
For maxima or minima, put $y^{\prime}=0$
$\Rightarrow \quad x=-\frac{1}{4}$
$y^{\prime \prime}=4=+\mathrm{ve}$
$y$ is minimum at $x=-\frac{1}{4}$.
Thus, minimum value $=2\left(-\frac{1}{4}\right)^2+\left(-\frac{1}{4}\right)-1$
$\frac{2}{16}-\frac{1}{4}-1=\frac{1}{8}-\frac{5}{4}=-\frac{9}{8}$
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