If 4+3x-7x^2 attains its maximum value M at x | vedclass

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(B) Let $f(x) = 4+3x-7x^2$. Taking the derivative, $f'(x) = 3-14x$. Setting $f'(x) = 0$, we get $x = \frac{3}{14} = \alpha$. Since $f''(x) = -14 < 0$,

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$23$

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(B) Let $f(x) = 4+3x-7x^2$. Taking the derivative, $f'(x) = 3-14x$. Setting $f'(x) = 0$, we get $x = \frac{3}{14} = \alpha$. Since $f''(x) = -14 The maximum value $M = f(\frac{3}{14}) = 4 + 3(\frac{3}{14}) - 7(\frac{3}{14})^2 = 4 + \frac{9}{14} - \frac{63}{196} = 4 + \frac{126-63}{196} = 4 + \frac{63}{196} = 4 + \frac{9}{28} = \frac{112+9}{28} = \frac{121}{28}$. Now, let $g(x) = 5x^2-2x+1$. Taking the derivative, $g'(x) = 10x-2$. Setting $g'(x) = 0$, we get $x = \frac{1}{5} = \beta$. Since $g''(x) = 10 > 0$, $g(x)$ attains its minimum at $x = \beta = \frac{1}{5}$. The minimum value $m = g(\frac{1}{5}) = 5(\frac{1}{5})^2 - 2(\frac{1}{5}) + 1 = \frac{1}{5} - \frac{2}{5} + 1 = \frac{4}{5}$. Finally, calculate $\frac{28(M-\alpha)}{5(m+\beta)} = \frac{28(\frac{121}{28} - \frac{3}{14})}{5(\frac{4}{5} + \frac{1}{5})} = \frac{28(\frac{121-6}{28})}{5(1)} = \frac{115}{5} = 23$.

If $4+3x-7x^2$ attains its maximum value $M$ at $x=\alpha$ and $5x^2-2x+1$ attains its minimum value $m$ at $x=\beta$, then $\frac{28(M-\alpha)}{5(m+\beta)}=$

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