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 the denominator,we get $y(x^2+3x+4) = x^2-3x+4$.Rearranging the terms,we get $(y-1)x

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

Vedclass · Answer

(C) Let $y = \frac{x^2-3x+4}{x^2+3x+4}$.Multiplying both sides by the denominator,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 $0$.$D = [3(y+1)]^2 - 4(y-1)(4(y-1)) \geq 0$.$9(y+1)^2 - 16(y-1)^2 \geq 0$.Expanding the squares: $9(y^2+2y+1) - 16(y^2-2y+1) \geq 0$.$9y^2 + 18y + 9 - 16y^2 + 32y - 16 \geq 0$.$-7y^2 + 50y - 7 \geq 0$.Multiplying by $-1$,we get $7y^2 - 50y + 7 \leq 0$.Factoring the quadratic: $(7y-1)(y-7) \leq 0$.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 $\frac{x^2-3x+4}{x^2+3x+4}$ will be

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