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

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(A) Let $y = \frac{x^2+14x+9}{x^2+2x+3}$.Since $x$ is real,the denominator $x^2+2x+3 = (x+1)^2+2$ is always positive.$y(x^2+2x+3) = x^2+14x+9$$x^2(y-1

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$4, -5$

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(A) Let $y = \frac{x^2+14x+9}{x^2+2x+3}$.Since $x$ is real,the denominator $x^2+2x+3 = (x+1)^2+2$ is always positive.$y(x^2+2x+3) = x^2+14x+9$$x^2(y-1) + 2x(y-7) + 3y-9 = 0$.For $x$ to be real,the discriminant $D \geq 0$.$D = [2(y-7)]^2 - 4(y-1)(3y-9) \geq 0$$4(y^2-14y+49) - 4(3y^2-12y+9) \geq 0$$y^2-14y+49 - 3y^2+12y-9 \geq 0$$-2y^2-2y+40 \geq 0$$y^2+y-20 \leq 0$$(y+5)(y-4) \leq 0$.Thus,$y \in [-5, 4]$.The maximum value is $4$ and the minimum value is $-5$.

If $x$ is real,then the maximum and minimum values of $\frac{x^2+14x+9}{x^2+2x+3}$ are respectively

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