Finding the range of a function that is defined for all real x How do I find the range of the function $ f(x)=\frac {x}{\sqrt {x^{2}+1}}$ in a practical way? Do I have to test values for x and see where the function is going?
With functions that are not defined for a particular x, I know there is going to be an horizontal asymptote, so I just find the limit where x is going towards infinity.
 A: Suppose $\;a\in\Bbb R\;$ is in the range, then there exists $\;x\in\Bbb R\;$ s.t.
$$f(x)=\frac x{\sqrt{x^2+1}}=a\implies x^2=a^2x^2+a^2\implies x^2=\frac{a^2}{1-a^2}$$
and the above means it must be $\;1-a^2>0\iff |a|<1\implies \;$ and thus $\;Im(f)\subset (-1,1)\;$ . Now, is the function onto this open interval? Perhaps one of the easiest ways to check this is by noting that
$$\lim_{x\to-\infty}f(x)=\frac x{-x\sqrt{1+\frac1{x^2}}}=-1\;,\;\;\lim_{x\to\infty}f(x)=\frac x{x\sqrt{1+\frac1{x^2}}}=1$$
and 
$$f'(x)=\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}=\frac1{{(x^2+1)^{3/2}}}>0\;\;\;\forall\,x\in\Bbb R$$
so $\;f\;$ is ascending monotonically and thus in fact $\;\text{Im}\,(f)=\left(-1,\,1\right)\;$
A: Hint: We have $\sqrt{x^2 + 1} > |x|$ for all $x \in \mathbb{R}$.

 Square both sides to obtain $x^2 + 1 > x^2$, which is clearly true. Therefore, $-1 < f(x) < 1$. 

Addendum: As previously mentioned in the comments by @KonKan this only shows that $\text{Im}(f) \subset (-1,1)$. To show the other inclusion one can (as many others have done so I won't repeat it) calculate $f'$ to see that $f$ is monotonically increasing globally and calculate $\lim_{x \to \pm \infty} f(x) = \pm 1$.
A: Hint:
WLOG $x=\tan y,-\dfrac\pi2<y<\dfrac\pi2\ \ \ \ (1)$ for finite real $x$
$\sec y=+\sqrt{x^2+1}$
$f(x)=\sin y$ where $y$ must honour $(1)$
Alternatively $$f^2(x)=1-\dfrac1{x^2+1}<1$$
$$\implies -1<f(x)<1$$
A: You can solve for $y$ the equation
$$
y=\frac{x}{\sqrt{x^2+1}}
$$
We can note that $y$ has the same sign as $x$ and square:
$$
y^2(x^2+1)=x^2
$$
that becomes $x^2(1-y^2)=y^2$ and therefore
$$
x=\frac{y}{\sqrt{1-y^2}}
$$
(we have no sign uncertainty because of the remark above). The right-hand side makes sense only for $-1<y<1$, so the range is $(-1,1)$.
Alternatively, note that
$$
-1<\frac{x}{\sqrt{x^2+1}}<1
$$
and that
$$
\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+1}}=-1,\qquad
\lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}=1
$$
Continuity and the intermediate value theorem tell you that the range is $(-1,1)$.
A: $f(x)=\dfrac{x}{\sqrt{x^2+1}}$.
$f(-x)=-f(x)$, the function is odd.
Let $x\ge 0$.
$f(0)=0$;
$0 \le f(x) =\dfrac{\sqrt{x^2}}{\sqrt{x^2+1}}<1$.
$\lim_{x \rightarrow \infty } f(x)=1$;
Since f is continuous (IVT):
For $x \ge 0$: Image $f=[0,1)$.
For  $x \in \mathbb {R}$ : Image $f =(-1,1)$ (why?)
(Recall $f$ is odd).
A: A precalculus approach: 
$$
y=\frac{x}{\sqrt{x^2+1}}\Leftrightarrow y^2(x^2+1)=x^2\Leftrightarrow (y^2-1)x^2+y^2=0
$$
You can view the last eq as a quadratic wrt to $x$ (given that $y\neq \pm 1$). Then its discriminant will be $D=-4y^2(y^2-1)$.  
But the range of $f$ consists of those values of $y$ for which there exist real solution(s) of the quadratic wrt $x$.  Then you should have $D\geq0$ i.e. $$y^2(y^2-1)\leq 0\Leftrightarrow y\in [-1,1]$$ 
and since $y\neq \pm 1$ we get that the acceptable values of $y$ (that is: the range if $f(x)$) are
$$
y\in(-1,1)
$$
A: First note that $$\frac{|x|}{\sqrt{x^2+1}}\le \frac{|x|}{\sqrt{x^2}}= \frac{|x|}{|x|}=1,$$ for all $x.$ Thus, $f(x)$ is bounded between $-1$ and $1.$ Also, we see that the limiting value of $f(x)$ as $x$ becomes $+\infty$ is $1.$ Since $f(x)$ is odd, it follows that $f(x)$ approaches $-1$ at $-\infty.$ From now on we can without loss of generality focus on the $+x$ axis.
Now since $f(x)$ is continuous you only need to see whether it ever attains the maximum value, $1.$ Thus setting $f(x)=1,$ you obtain $x=\sqrt{x^2+1},$ or $$x^2=x^2+1.$$ This is never true. Thus, your range is $(-1,1).$
