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

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(C) Let $y = \frac{x^2-3x+4}{x^2+3x+4}$.$y(x^2+3x+4) = x^2-3x+4$$(y-1)x^2 + 3(y+1)x + 4(y-1) = 0$.Since $x$ is real,the discriminant $D \geq 0$.$D = [

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $y = \frac{x^2-3x+4}{x^2+3x+4}$.$y(x^2+3x+4) = x^2-3x+4$$(y-1)x^2 + 3(y+1)x + 4(y-1) = 0$.Since $x$ is real,the discriminant $D \geq 0$.$D = [3(y+1)]^2 - 4(y-1)(4(y-1)) \geq 0$$9(y+1)^2 - 16(y-1)^2 \geq 0$$9(y^2+2y+1) - 16(y^2-2y+1) \geq 0$$-7y^2 + 50y - 7 \geq 0$$7y^2 - 50y + 7 \leq 0$$(7y-1)(y-7) \leq 0$.This inequality holds when $\frac{1}{7} \leq y \leq 7$.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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