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(A) Since $f$ is an even function,$f(-x) = f(x)$ for all $x \in (-5, 5)$.Given $f(x) = f\left( \frac{x + 1}{x + 2} \right)$.Since $f(x) = f(-x)$,we ha

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$\frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2}$

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(A) Since $f$ is an even function,$f(-x) = f(x)$ for all $x \in (-5, 5)$.Given $f(x) = f\left( \frac{x + 1}{x + 2} \right)$.Since $f(x) = f(-x)$,we have $f(x) = f\left( \frac{-x + 1}{-x + 2} \right)$.Case $1$: $x = \frac{-x + 1}{-x + 2} \Rightarrow -x^2 + 2x = -x + 1 \Rightarrow x^2 - 3x + 1 = 0$.Solving for $x$,we get $x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$.Case $2$: $-x = \frac{x + 1}{x + 2} \Rightarrow -x^2 - 2x = x + 1 \Rightarrow x^2 + 3x + 1 = 0$.Solving for $x$,we get $x = \frac{-3 \pm \sqrt{9 - 4}}{2} = \frac{-3 \pm \sqrt{5}}{2}$.Thus,the four values are $\frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2}$.

If $f$ is an even function defined on the interval $(-5, 5)$,then four real values of $x$ satisfying the equation $f(x) = f\left( \frac{x + 1}{x + 2} \right)$ are

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