# Find the difference of maximum and minimum values of x^+4x+9 / x^2 +9

Solution:

$\\ \\ \Rightarrow \hspace{1cm} y=\frac{x^{2}+4x+9}{x^{2}+9}\\ \\ \Rightarrow \hspace{1cm}(y-1)x^{2}-4x+9(y-1)=0\\ \\ \Rightarrow \hspace{1cm}D\geq0\Rightarrow 16-36(y-1)^{2}\geq0\\ \\ \Rightarrow \hspace{1cm}4-9(y-1)^{2}\geq0\\ \\ \Rightarrow \hspace{1cm}\frac{1}{3}\leq y\leq \frac{5}{3}$

$\therefore$     Difference of maximum and minimum values is

$\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$

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