If x is real,then the maximum and minimum val | vedclass

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(C) Let $y = \frac{x^2 - 3x + 4}{x^2 + 3x + 4}$.Multiplying both sides by $(x^2 + 3x + 4)$,we get $y(x^2 + 3x + 4) = x^2 - 3x + 4$.Rearranging the ter

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$7, \frac{1}{7}$

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(C) Let $y = \frac{x^2 - 3x + 4}{x^2 + 3x + 4}$.Multiplying both sides by $(x^2 + 3x + 4)$,we get $y(x^2 + 3x + 4) = x^2 - 3x + 4$.Rearranging the terms,we get $(y - 1)x^2 + 3(y + 1)x + 4(y - 1) = 0$.Since $x$ is real,the discriminant $D$ must be greater than or equal to $0$.$D = [3(y + 1)]^2 - 4(y - 1)(4(y - 1)) \ge 0$.$9(y + 1)^2 - 16(y - 1)^2 \ge 0$.$(3(y + 1))^2 - (4(y - 1))^2 \ge 0$.Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $(3y + 3 - 4y + 4)(3y + 3 + 4y - 4) \ge 0$.$(-y + 7)(7y - 1) \ge 0$.Multiplying by $-1$,we get $(y - 7)(7y - 1) \le 0$.This inequality holds when $\frac{1}{7} \le y \le 7$.Thus,the maximum value is $7$ and the minimum value is $\frac{1}{7}$.

If $x$ is real,then the maximum and minimum values of the expression $\frac{x^2 - 3x + 4}{x^2 + 3x + 4}$ are

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