The minimum value of 2x^2+x-1 is: | vedclass

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(C) Let $y = 2x^2 + x - 1$. To find the minimum value,we differentiate $y$ with respect to $x$: $y' = 4x + 1$. For critical points,set $y' = 0$: $4x +

Vedclass · Accepted Answer

$-\frac{9}{8}$

Vedclass · Answer

(C) Let $y = 2x^2 + x - 1$. To find the minimum value,we differentiate $y$ with respect to $x$: $y' = 4x + 1$. For critical points,set $y' = 0$: $4x + 1 = 0 \Rightarrow x = -\frac{1}{4}$. Now,find the second derivative: $y'' = 4$. Since $y'' > 0$,the function has a minimum at $x = -\frac{1}{4}$. Substituting $x = -\frac{1}{4}$ into the original equation: $y = 2(-\frac{1}{4})^2 + (-\frac{1}{4}) - 1$ $y = 2(\frac{1}{16}) - \frac{1}{4} - 1$ $y = \frac{1}{8} - \frac{2}{8} - \frac{8}{8} = -\frac{9}{8}$. Thus,the minimum value is $-\frac{9}{8}$.

The minimum value of $2x^2+x-1$ is:

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