For real x, the greatest value of frac{x^{2}+ | vedclass

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(C) Let $y = \frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$.$y(2 x^{2}+4 x+9) = x^{2}+2 x+4$.$2 y x^{2} + 4 y x + 9 y = x^{2} + 2 x + 4$.$(2 y-1) x^{2} + (4 y-2)

Vedclass · Accepted Answer

$\frac{1}{2}$

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(C) Let $y = \frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$.$y(2 x^{2}+4 x+9) = x^{2}+2 x+4$.$2 y x^{2} + 4 y x + 9 y = x^{2} + 2 x + 4$.$(2 y-1) x^{2} + (4 y-2) x + (9 y-4) = 0$.Since $x$ is a real number, the discriminant $D \geq 0$.$D = (4 y-2)^{2} - 4(2 y-1)(9 y-4) \geq 0$.$4(2 y-1)^{2} - 4(2 y-1)(9 y-4) \geq 0$.$4(2 y-1) [ (2 y-1) - (9 y-4) ] \geq 0$.$4(2 y-1) (-7 y + 3) \geq 0$.$(2 y-1) (7 y - 3) \leq 0$.This inequality holds for $\frac{3}{7} \leq y \leq \frac{1}{2}$.Therefore, the maximum value of $y$ is $\frac{1}{2}$.

For real $x$, the greatest value of $\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$ is

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