The range of the function f(x) = frac{x}{x^2 | vedclass

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(B) Given,$f(x) = \frac{x}{x^2 - 5x + 9} = y$. Since $x^2 - 5x + 9$ has a discriminant $D = (-5)^2 - 4(1)(9) = 25 - 36 = -11 < 0$,the denominator is a

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$\left[-\frac{1}{11}, 1\right]$

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(B) Given,$f(x) = \frac{x}{x^2 - 5x + 9} = y$. Since $x^2 - 5x + 9$ has a discriminant $D = (-5)^2 - 4(1)(9) = 25 - 36 = -11 Rearranging the equation: $y(x^2 - 5x + 9) = x \Rightarrow yx^2 - (5y + 1)x + 9y = 0$. For $x$ to be a real number,the discriminant of this quadratic equation must be greater than or equal to zero $(D \geq 0)$. $D = (-(5y + 1))^2 - 4(y)(9y) \geq 0$ $(5y + 1)^2 - 36y^2 \geq 0$ $25y^2 + 10y + 1 - 36y^2 \geq 0$ $-11y^2 + 10y + 1 \geq 0$ $11y^2 - 10y - 1 \leq 0$ Factoring the quadratic: $(11y + 1)(y - 1) \leq 0$. The critical points are $y = -\frac{1}{11}$ and $y = 1$. Testing the intervals,the inequality holds for $y \in \left[-\frac{1}{11}, 1\right]$.

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

The range of the function $f(x) = \frac{x}{x^2 - 5x + 9}$ is

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