What is the maximum value of the expression f | vedclass

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Question

(A) Let $y = \frac{x}{x^2 - 5x + 9}$.Then $y(x^2 - 5x + 9) = x$,which implies $yx^2 - (5y + 1)x + 9y = 0$.Since $x$ is a real number,the discriminant

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $y = \frac{x}{x^2 - 5x + 9}$.Then $y(x^2 - 5x + 9) = x$,which implies $yx^2 - (5y + 1)x + 9y = 0$.Since $x$ is a real number,the discriminant $D \ge 0$.$D = (-(5y + 1))^2 - 4(y)(9y) \ge 0$.$(5y + 1)^2 - 36y^2 \ge 0$.$25y^2 + 10y + 1 - 36y^2 \ge 0$.$-11y^2 + 10y + 1 \ge 0$.$11y^2 - 10y - 1 \le 0$.$(11y + 1)(y - 1) \le 0$.This inequality holds for $y \in [-\frac{1}{11}, 1]$.Thus,the maximum value is $1$.

What is the maximum value of the expression $\frac{x}{x^2 - 5x + 9}$ for all real values of $x$?

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