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

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Question

(a) Since $f$ is an even function $f\,( - x) = f(x)\,,\forall x \in ( - 5,\,5).$

We are given that $f(x) = f\,\left( {\frac{{x + 1}}{{x + 2}}} \rig

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) Since $f$ is an even function $f\,( - x) = f(x)\,,\forall x \in ( - 5,\,5).$

We are given that $f(x) = f\,\left( {\frac{{x + 1}}{{x + 2}}} \right)$

$ \Rightarrow \,\,f( - x) = f\left( {\frac{{ - x + 1}}{{ - x + 2}}} \right)\, \Rightarrow \,\,f(x) = f\,\left( {\frac{{ - x + 1}}{{ - x + 2}}} \right)$

To find the values of $x$, we set

$x = \frac{{ - x + 1}}{{ - x + 2}}\,\, \Rightarrow \,\,x = \frac{{3 \pm \sqrt {9 - 4} }}{2} = \frac{{3 \pm \sqrt 5 }}{2}$

Also $f(x) = f\,\left( {\frac{{x + 1}}{{x + 2}}} \right) = f( - x)$

To find the values of $x$, we set

$ - x = \frac{{x + 1}}{{x + 2}}\,\, \Rightarrow \,\,x = \frac{{ - 3 \pm \sqrt {9 - 4} }}{2} = \frac{{ - 3 \pm \sqrt 5 }}{2}$

Thus the four required values of $x$ 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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