How to find range of $\frac{\sqrt{1+2x^2}}{1+x^2}$? 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.
 A: 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 $$\dfrac{\sqrt{1+2x^2}}{1+x^2}=\dfrac{2\tan v}{1+\tan^2v}=\sin2v$$
A: \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}
$$\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{\sqrt{1+2{{x}^{2}}}}{1+{{x}^{2}}}=0$$
$$R_f=(0,1]$$

A: 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: $y\in(0,1]$.
