Question

find the difference of maximum and minimum value of x^2+4x+9÷x^2+9​

Answer 1

Answer:

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Answer 2

Answer:

The difference between maximum and minimum values is$\frac{4}{3}$

Step-by-step explanation:

$\begin{array}{ll}\Rightarrow & y=\frac{x^{2}+4 x+9}{x^{2}+9} \\\Rightarrow & (y-1) x^{2}-4 x+9(y-1)=0 \\\Rightarrow & D \geq 0 \Rightarrow 16-36(y-1)^{2} \geq 0 \\\Rightarrow & 4-9(y-1)^{2} \geq 0 \\\Rightarrow & \frac{1}{3} \leq y \leq \frac{5}{3}\end{array}$

$\therefore$ The difference between maximum and minimum values is

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