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If $g$ is an even function defined in $(-5,5)$ . Then number of real $x$ in $$g\bigg(\frac{x+1}{x+2}\bigg) = g(x)$$

Attempt: for even function , $g(-x) = g(x)$

So $$g\bigg(\frac{x+1}{x+2}\bigg) = g(x) = g(-x)$$

So $$\frac{x+1}{x+2} = x\Rightarrow x^2+x-1=0\Rightarrow x= \frac{-1\pm \sqrt{5}}{2}$$ or $$\frac{x+1}{x+2} = -x\Rightarrow x^2+3x+1=0\Rightarrow x=\frac{-3\pm \sqrt{5}}{2}$$

But answer given as $$x=\frac{-3\pm \sqrt{5}}{2}\;, \frac{3\pm \sqrt{5}}{2}$$

could some help me how to get $\displaystyle \frac{3\pm \sqrt{5}}{2}$ and what,s wrong in my attempt, thanks

DXT
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    This does not look quite right to me. $g(x)=g(y)$ cannot have $x=y$ or $x=-y$. One counter example $g(x)=\cos 5\pi x$ – Jay Zha May 03 '17 at 03:25
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    Then number of real x That depends on $g,$. For example, $g$ could be a constant function, then the equation would have infinitely many solutions. – dxiv May 03 '17 at 03:25
  • Thanks dxiv. same question asked here https://math.stackexchange.com/questions/1742611/if-f-is-an-even-function-defined-on-the-interval-5-5-then-four-real-value?rq=1 – DXT May 03 '17 at 03:28
  • please explain me what should be the minimum number of real $x$ in above equation, thanks – DXT May 03 '17 at 03:29
  • It was too long for a comment, so I posted it as an answer. You missed some cases, otherwise you did nothing wrong. As for why the given answer excluded the $\frac{1}{2}(\pm1 \pm\sqrt{5})$ maybe there was some additional condition, or maybe something got lost in translation. Note that the original question that you linked asked for only four real values of x satisfying the equation, it didn't say that there couldn't be more. – dxiv May 03 '17 at 03:58

1 Answers1

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Since $\;g\left(\cfrac{x+1}{x+2}\right) = g(x) = g(-x) = g\left(\cfrac{-x+1}{-x+2}\right)\,$ all of the following that fall within $(-5,5)$ will qualify as solutions (which, as it happens, they all do):

  • $\;\cfrac{x+1}{x+2} = x \;\;\iff\;\; x^2+x-1=0 \;\;\iff\;\; x = \cfrac{-1 \pm \sqrt{5}}{2}$

  • $\;\cfrac{x+1}{x+2} = -x \;\;\iff\;\; x^2+3x+1=0 \;\;\iff\;\; x = \cfrac{-3 \pm \sqrt{5}}{2}$

  • $\;\cfrac{-x+1}{-x+2} = x \;\;\iff\;\; x^2-3x+1=0 \;\;\iff\;\; x = \cfrac{3 \pm \sqrt{5}}{2}$

  • $\;\cfrac{-x+1}{-x+2} = -x \;\;\iff\;\; x^2-x-1=0 \;\;\iff\;\; x = \cfrac{1 \pm \sqrt{5}}{2}$

dxiv
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  • Could you help me with.this ques. https://math.stackexchange.com/questions/2263164/consider-a-right-angled-triangle-pqr-right-angled-at-p-i-e-angle-qpr-90 – pi-π May 03 '17 at 09:02