If the extreme value of 3x - 2x^2 + 1 is k, t | vedclass

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(C) Let $f(x) = -2x^2 + 3x + 1$. To find the extreme value, we differentiate $f(x)$ with respect to $x$: $f'(x) = -4x + 3$. Setting $f'(x) = 0$, we ge

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$(-\infty, \infty)$

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(C) Let $f(x) = -2x^2 + 3x + 1$. To find the extreme value, we differentiate $f(x)$ with respect to $x$: $f'(x) = -4x + 3$. Setting $f'(x) = 0$, we get $-4x + 3 = 0$, which implies $x = \frac{3}{4}$. The extreme value $k$ is $f\left(\frac{3}{4}\right) = -2\left(\frac{3}{4}\right)^2 + 3\left(\frac{3}{4}\right) + 1 = -2\left(\frac{9}{16}\right) + \frac{9}{4} + 1 = -\frac{9}{8} + \frac{18}{8} + \frac{8}{8} = \frac{17}{8}$. Now, we need to find the set of $x$ such that $kx^2 + 2x + 1 > 0$, where $k = \frac{17}{8}$. Substituting $k$, we get $\frac{17}{8}x^2 + 2x + 1 > 0$. Multiplying by $8$, we get $17x^2 + 16x + 8 > 0$. The discriminant $D$ of this quadratic is $D = b^2 - 4ac = (16)^2 - 4(17)(8) = 256 - 544 = -288$. Since $D Thus, the solution set is $(-\infty, \infty)$.

If the extreme value of $3x - 2x^2 + 1$ is $k$, then the set of all real values of $x$ for which $kx^2 + 2x + 1 > 0$ is

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