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

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Question

(C) Let $y = 2x^2 + x - 1$. To find the minimum value, we find the derivative $y' = 4x + 1$. Setting $y' = 0$ for critical points, we get $4x + 1 = 0$

Vedclass · Accepted Answer

$-\frac{9}{8}$

Vedclass · Answer

(C) Let $y = 2x^2 + x - 1$. To find the minimum value, we find the derivative $y' = 4x + 1$. Setting $y' = 0$ for critical points, we get $4x + 1 = 0$, which implies $x = -\frac{1}{4}$. The second derivative is $y'' = 4$, which is positive $(> 0)$, confirming that the function has a minimum at $x = -\frac{1}{4}$. Substituting $x = -\frac{1}{4}$ into the original expression: $y = 2(-\frac{1}{4})^2 + (-\frac{1}{4}) - 1$ $y = 2(\frac{1}{16}) - \frac{1}{4} - 1$ $y = \frac{1}{8} - \frac{2}{8} - \frac{8}{8} = -\frac{9}{8}$. Thus, the minimum value is $-\frac{9}{8}$.

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

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