Find the image and inverse image of the following function $2x - x^2$ Where $f(E)$ denotes the image and $f^{-1}(E)$ denotes the inverse image.
 $$E=[-2,2)$$
plugging in the values I get:
$f(-2)=-8$
$f(-1)= -3$
$f(0) = 0$
$f(1)= 1$
$f(2) = 0$
So all the numbers fall between the interval of $[-8,1)$ I think it has to be expressed differently than $[-8,1)$ because 0 appears twice. How do I find the inverse image? I cannot algebraically solve for the inverse function. 
Edit: Looking at the graph on desmos APPEARS to verify my interval for $y$. Can I 

We can complete the square to determine the image.
\begin{align*}
f(x) & = 2x - x^2\\
     & = -x^2 + 2x\\
     & = -(x^2 - 2x)\\
     & = -(x^2 - 2x + 1) + 1\\
     & = -(x - 1)^2 + 1
\end{align*}
Thus, the graph of $f$ is a parabola with vertex $(1, 1)$ that opens downwards.  Therefore, its maximum value on the interval $E = [-2, 2)$ is $1$.  The minimum value must occur at the endpoint $x = -2$ since $-2$ is farther from the axis of symmetry $x = 1$ of the function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = 2x - x^2$ than $x = 2$ is.  Since $f(-2) = -8$, the range of the continuous function $f: [-2, 2) \to \mathbb{R}$ defined by $f(x) = 2x - x^2 = -(x - 1)^2 + 1$ is $[-8, 1]$, as the graph below confirms.  

The formal justification for the assertion that $f$ assumes every value between $-8$ and $1$ is the Intermediate Value Theorem, which states that if a function is continuous on an interval $[a, b]$, then it takes on every value between $f(a)$ and $f(b)$.  In this case, we can apply the Intermediate Value Theorem to the interval $[-2, 1] \subseteq [-2, 2)$ to conclude that $f$ assumes every value between $f(-2) = -8$ and $f(1) = 1$.
As for $f^{-1}(E)$, we require that $-2 \leq f(x) < 2$.
\begin{align*}
-2 & \leq f(x) < 2\\
-2 & \leq -(x - 1)^2 + 1 < 2\\
2 & \geq (x - 1)^2 - 1 > -2\\
3 & \geq (x - 1)^2 > -3
\end{align*}
As Kavi Rama Murthy explained, the inequality $(x - 1)^2 > -3$ is automatically satisfied since $(x - 1)^2 \geq 0$ for any real number $x$.  Hence, 
\begin{align*}
(x - 1)^2 & \leq 3\\
|x - 1| & \leq \sqrt{3} && \text{since $\sqrt{u^2} = |u|$}
\end{align*}
Since $|x - 1| \leq \sqrt{3}$ means $x - 1$ is at most $\sqrt{3}$ units from $0$, 
\begin{align*}
-\sqrt{3} \leq x - 1 \leq \sqrt{3}\\
1 - \sqrt{3} \leq x \leq 1 + \sqrt{3}
\end{align*}
Thus, $f^{-1}(E) = [-1 - \sqrt{3}, 1 + \sqrt{3}]$.
