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How to find range of $$\frac{\sqrt{1+2x^2}}{1+x^2}$$ ?

I tried put it equal to $y$ and squaring but I'm getting $4$th degree equation.

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  • $\begingroup$ Hint: Note that you cannot get negative values and you can only approach zero as $x\to\pm\infty$. So the only question is how large it can be. $\endgroup$ – almagest Jun 20 '16 at 8:00
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    $\begingroup$ This is also $$\sqrt{1-\frac{x^4}{(1+x^2)^2}}=\sqrt{1-\frac{1}{(1+1/x^2)^2}}.$$ Now, the range of $1+1/x^2$ is $____$ hence the range of $1/(1+1/x^2)^2$ is $____$ hence the range of $1-1/(1+1/x^2)^2$ is $____$ and finally the desired range is $____$. $\endgroup$ – Did Jun 20 '16 at 8:36
  • $\begingroup$ @Did Nice.You might consider making it an answer :-)! $\endgroup$ – user220382 Jun 20 '16 at 9:14
  • $\begingroup$ Let me suggest instead that you develop these indications into a full answer and that you post it below yourself. After a while, you may even want to accept it. $\endgroup$ – Did Jun 20 '16 at 9:22
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Let $\sqrt{1+2x^2}=u\ge1,\implies1+x^2=\dfrac{1+u^2}2$

$$\dfrac{\sqrt{1+2x^2}}{1+x^2}=\dfrac{2u}{1+u^2}=\dfrac2{u+\dfrac1u}$$

Now $u+\dfrac1u\ge2\sqrt{u\cdot\dfrac1u}=2$


Alternatively, let $\sqrt{1+2x^2}=\tan v$

Clearly, $\tan v\ge1+2\cdot0=1$ WLOG we can choose $\dfrac\pi4\le v<\dfrac\pi2\iff\dfrac\pi2\le2v<\pi$

Now $$\dfrac{\sqrt{1+2x^2}}{1+x^2}=\dfrac{2\tan v}{1+\tan^2v}=\sin2v$$

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  • $\begingroup$ Brilliant. _/_ How did you get that idea :-P ? $\endgroup$ – user220382 Jun 20 '16 at 8:01
  • $\begingroup$ @SanchayanDutta, Tried to rationalize :) $\endgroup$ – lab bhattacharjee Jun 20 '16 at 8:01
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    $\begingroup$ @SanchayanDutta, Please find the alternative approach too:) It gives us directly both the max, min $\endgroup$ – lab bhattacharjee Jun 20 '16 at 8:05
  • $\begingroup$ @SanchayanDutta, Btw, Have you noticed : math.stackexchange.com/questions/1817300/… $\endgroup$ – lab bhattacharjee Jun 20 '16 at 8:12
  • $\begingroup$ Oh i missed it.Thanks.Seeing right away ! $\endgroup$ – user220382 Jun 20 '16 at 8:18
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\begin{align} & \frac{\sqrt{1+2{{x}^{2}}}}{1+{{x}^{2}}}>0 \\ & f'(x)=\frac{-2{{x}^{3}}}{{{(1+{{x}^{2}})}^{2}}\sqrt{1+2{{x}^{2}}}} \\ & f(0)=1 \\ \end{align} $$\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{\sqrt{1+2{{x}^{2}}}}{1+{{x}^{2}}}=0$$ $$R_f=(0,1]$$ enter image description here

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You indeed get a fourth degree equation, but $x$ only appears with even exponent: $$ y=\frac{\sqrt{1+2x^2}}{1+x^2} $$ means that $y>0$ and that $$ (1+x^2)^2y^2=1+2x^2 $$ Expanding and reordering gives $$ y^2x^4+2(y^2-1)x^2+y^2-1=0 $$ and the usual quadratic formula provides the value for $x^2$; it's common to advise setting $z=x^2$ and solving $y^2z^2+2(y^2-1)z+y^2-1=0$: $$ z=\frac{1-y^2\pm\sqrt{1-y^2}}{y^2} $$ Note that the discriminant should be nonnegative, thus $y^2\le1$. In this case one of the roots of the quadratic is negative and the other one is positive (use Descartes' rule of signs). Hence, for $y^2\le1$, the equation $$ x^2=\frac{1-y^2+\sqrt{1-y^2}}{y^2} $$ has two solutions (except for $y=1$).

Now the conditions $$ \begin{cases} y>0 \\[4px] y^2\le1 \end{cases} $$ give you the range: $y\in(0,1]$.

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