What is the range of $\alpha$? 
My trial was through graphical method .
I plotted the function varying $\alpha$. As a result a void was formed in the region when $\alpha$ belongs to 
$R \setminus [-2,2]$. Is there any explanation for that by algebraic method or by calculus.  I tried finding minima, maximal but couldn't arrive at the range of $\alpha$. Any hints to arrive at the solution will be appreciated. Thanks in advance.
 A: Start by plotting the function
$$g(x) = x^3 - 3x + 6.$$
This function is identical to $f$, except the natural domain of $f$ doesn't contain the point $\alpha$. Basically, there will be a hole in the graph where $x = \alpha$, but otherwise the graphs are the same. Depending on where that hole is, this could remove a point from the range of $f$.
The function $g$ has a range of $\Bbb{R}$, but it is not one-to-one. Most values in $\Bbb{R}$ are achieved once by $g$, but some are achieved three times (and a couple are achieved twice). If $g(\alpha)$ is only achieved once (i.e. at $x = \alpha$), then removing this point from the graph will make the range of $f$ not equal to $\Bbb{R}$. Otherwise, if $g(\alpha)$ is achieved elsewhere, then removing $x = \alpha$ will not affect the range at all, and hence the range of $f$ would be $\Bbb{R}$.
Let's start by computing the stationary points of $g$. We have
$$0 = g'(x) = 3x^2 - 3 \implies x = \pm 1.$$
We get a local maximum at $x = -1$, at which point we have $g(-1) = 8$. We also get a local minimum at $x = -1$, at which point $g(1) = 4$. So, looking at the cubic graph, every value between $4$ and $8$ (inclusive) is duplicated at least once. So long as $g(\alpha) \notin [4, 8]$, then $f$ will not have full range.
So, we solve $g(\alpha) < 4$ or $g(\alpha) > 8$. We have $g(\alpha) < 4$ if and only if
\begin{align*}
\alpha^3 - 3\alpha + 6 < 4 &\iff \alpha^3 - 3\alpha + 2 < 0 \\
&\iff (\alpha - 1)^2(\alpha + 2) < 0 \\
&\iff \alpha < -2 \text{ and } \alpha \neq 1 \\
&\iff \alpha < -2.
\end{align*}
On the other hand, we have $g(\alpha) > 8$ if and only if
\begin{align*}
\alpha^3 - 3\alpha + 6 > 8 &\iff \alpha^3 - 3\alpha - 2 > 0 \\
&\iff (\alpha + 1)^2(\alpha - 2) > 0 \\
&\iff \alpha > 2 \text{ and } \alpha \neq -1 \\
&\iff \alpha > 2.
\end{align*}
Thus, $f$ does not have full range, if and only if $\alpha \notin [-2, 2]$, i.e. $\alpha \in \Bbb{R} - [-2, 2]$.
A: The domain of this function is $x \neq \alpha$, as $x-\alpha$ appears at the denominator. For $x \neq \alpha$ you can simplify the expression to $f(x) = (1/2)(x^3-3x+6)$, which is a cubic defined for all $x$ but $\alpha$. Without necessarily plotting the function, you can rearrange the terms so that 
\begin{align*}
f(x) & = \frac12(x^3-3x+6)\\
& = \frac12x(x^2-3)+3.
\end{align*}
This tells you that $f(x)-3$ is an odd cubic with three intersections with the $x$-axis, namely $0,\pm \sqrt3$. You have a local maximum at $x=-1$ and a local minimum at $x=1$. Since $f(x)-3$ is odd you can compute $f(1) = 2$ and automatically get $f(-1)=4$. Now the crucial observation is that if $f(\alpha)$ is between $2$ and $4$, then the range of $f$ is still all of $\mathbb{R}$. This happens when $\alpha \in [-2,2]$, as you can see from computing $x$ such that $f(x) = 2$ and $f(x) = 4$. So the solution is $\mathbb{R} \setminus [-2,2]$.
