If x is real,the expression frac{x + 2}{2x^2 | vedclass

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(B) Let $y = \frac{x + 2}{2x^2 + 3x + 6}$.Then $y(2x^2 + 3x + 6) = x + 2$,which simplifies to $2yx^2 + (3y - 1)x + (6y - 2) = 0$.If $y = 0$,then $x =

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$\left[ -\frac{1}{13}, \frac{1}{3} \right]$

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(B) Let $y = \frac{x + 2}{2x^2 + 3x + 6}$.Then $y(2x^2 + 3x + 6) = x + 2$,which simplifies to $2yx^2 + (3y - 1)x + (6y - 2) = 0$.If $y = 0$,then $x = -2$,which is a real value.If $y 
eq 0$,for $x$ to be real,the discriminant $D \ge 0$.$D = (3y - 1)^2 - 4(2y)(6y - 2) \ge 0$.$9y^2 - 6y + 1 - 48y^2 + 16y \ge 0$.$-39y^2 + 10y + 1 \ge 0$.$39y^2 - 10y - 1 \le 0$.Factoring the quadratic: $(13y + 1)(3y - 1) \le 0$.This inequality holds for $y \in \left[ -\frac{1}{13}, \frac{1}{3} ight]$.Since $y = 0$ is included in this interval,the range is $\left[ -\frac{1}{13}, \frac{1}{3} ight]$.

If $x$ is real,the expression $\frac{x + 2}{2x^2 + 3x + 6}$ takes all values in the interval

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