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

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(C) Let $f(x) = 2x^2 + x - 1$.To find the minimum value,we find the derivative $f'(x)$ and set it to $0$:$f'(x) = 4x + 1$.Setting $f'(x) = 0$,we get $

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$-\frac{9}{8}$

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(C) Let $f(x) = 2x^2 + x - 1$.To find the minimum value,we find the derivative $f'(x)$ and set it to $0$:$f'(x) = 4x + 1$.Setting $f'(x) = 0$,we get $4x + 1 = 0$,which implies $x = -\frac{1}{4}$.Now,we check the second derivative $f''(x) = 4$. Since $f''(x) > 0$,the function has a local minimum at $x = -\frac{1}{4}$.The minimum value is $f(-\frac{1}{4}) = 2(-\frac{1}{4})^2 + (-\frac{1}{4}) - 1$.$f(-\frac{1}{4}) = 2(\frac{1}{16}) - \frac{1}{4} - 1 = \frac{1}{8} - \frac{2}{8} - \frac{8}{8} = -\frac{9}{8}$.

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

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