Question

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Answer 1

Step-by-step explanation:

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Answer 2

Solution

Suppose a line $y=k$ to intersect $y=\dfrac{x}{x^2+5x+9}$.

$\implies \dfrac{x}{x^2+5x+9} =k$

$\implies x=k(x^2+5x+9)$

$\implies kx^2+(5k-1)x+9k=0$

If the graph of $y=\dfrac{x}{x^2+5x+9}$ and $y=k$ intersect, there are real solutions to the equation $kx^2+(5k-1)x+9k=0$.

$\implies D=(5k-1)^2-4(k)(9k)=25k^2-10k+1-36k^2=\boxed{-11k^2-10k+1}$

$\implies -11k^2-10k+1\geq 0$

$\implies 11k^2+10k-1\leq 0$

$\implies (11k+1)(k-1)\leq 0$

The solution to this inequality is $\boxed{-\dfrac{1}{11} \leq k\leq 1}$. Hence, the maximum value and the minimum value of $\dfrac{x}{x^2+5x+9}$ is $\boxed{-\dfrac{1}{11}}$ and $\boxed{1}$ respectively.