left{x in R : frac{sqrt{6+x-x^2}}{2x+5} geq f | vedclass

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(C) The given inequality is $\frac{\sqrt{6+x-x^2}}{2x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}$.First,for the expression to be defined,we must have $6+x-x^2

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$[-2,-1] \cup \{3\}$

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(C) The given inequality is $\frac{\sqrt{6+x-x^2}}{2x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}$.First,for the expression to be defined,we must have $6+x-x^2 \geq 0$,which implies $x^2-x-6 \leq 0$,so $(x-3)(x+2) \leq 0$. Thus,$x \in [-2, 3]$.If $6+x-x^2 = 0$,then $x = -2$ or $x = 3$. Both satisfy the inequality as $0 \geq 0$.If $6+x-x^2 > 0$,we can divide by $\sqrt{6+x-x^2}$:$\frac{1}{2x+5} \geq \frac{1}{x+4} \Rightarrow \frac{1}{2x+5} - \frac{1}{x+4} \geq 0$$\Rightarrow \frac{x+4-(2x+5)}{(2x+5)(x+4)} \geq 0$ $\Rightarrow \frac{-x-1}{(2x+5)(x+4)} \geq 0$$\Rightarrow \frac{x+1}{(2x+5)(x+4)} \leq 0$.Using the sign scheme for critical points $-4, -2.5, -1$,the expression is $\leq 0$ for $x \in (-\infty, -4) \cup (-2.5, -1]$.Combining this with the domain $x \in [-2, 3]$,we get $x \in [-2, -1] \cup \{3\}$.

$\left\{x \in R : \frac{\sqrt{6+x-x^2}}{2x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}\right\}=$

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