If f is an even function defined on the inter | vedclass

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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 $f(x) = f(-x)$,the e

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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 $f(x) = f(-x)$,the equation $f(x) = f\left( \frac{x + 1}{x + 2} \right)$ is satisfied if $x = \frac{x + 1}{x + 2}$ or if $-x = \frac{x + 1}{x + 2}.$Case $1$: $x = \frac{x + 1}{x + 2} \Rightarrow x^2 + 2x = x + 1 \Rightarrow x^2 + x - 1 = 0.$Using the quadratic formula,$x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}.$Case $2$: $-x = \frac{x + 1}{x + 2} \Rightarrow -x^2 - 2x = x + 1 \Rightarrow x^2 + 3x + 1 = 0.$Using the quadratic formula,$x = \frac{-3 \pm \sqrt{9 - 4(1)(1)}}{2} = \frac{-3 \pm \sqrt{5}}{2}.$Thus,the four values are $\frac{-1 \pm \sqrt{5}}{2}$ and $\frac{-3 \pm \sqrt{5}}{2}.$ Note: The provided options suggest the intended roots are $\frac{3 \pm \sqrt{5}}{2}$ and $\frac{-3 \pm \sqrt{5}}{2}$ based on the equation $f(x) = f\left( \frac{-x+1}{-x+2} \right)$ logic.

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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