The range of the function $f(x) = \frac{2}{2+4x^2-4x}$ can be written as an interval $(a,b]$. What is $a+b$?
For $f(x)$ to take on all real values, the denominator $2+4x^2-4x$ cannot be equal to zero. This is because division by zero is undefined, so we need to exclude values of $x$ that make the denominator zero.
We have the quadratic $2+4x^2-4x=0$. Moving all terms to one side gives $4x^2-4x+2=0$. We cannot factor this quadratic, so we apply the quadratic formula: \begin{align*}
x&=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(4)(2)}}{2(4)}\\
&=\dfrac{4\pm\sqrt{16-32}}{8}\\
&=\dfrac{4\pm\sqrt{-16}}{8}=\dfrac{4\pm4i}{8}=\dfrac{1\pm i}{2}.
\end{align*}Therefore, $(2+4x^2-4x)=(x-(1+i))(x-(1-i))$. Hence, the denominator can be factored as $2+4x^2-4x=2(x-(1+i))(x-(1-i))$, and it follows that \begin{align*}
f(x)&=\frac{2}{2(x-(1+i))(x-(1-i))}\\
&=\frac{2}{2(x-1-i)(x-1+i)}=\frac{1}{(x-1-i)(x-1+i)}.
\end{align*}
Now, notice that for any complex number $z=a+bi$, \[|z|^2=z\cdot\overline{z}=(a+bi)(a-bi)=a^2+b^2.\]This means that $|z|$ represents the distance from $z$ to the origin.
Thus, if $x$ is real, $x-(1+i)$ and $x-(1-i)$ are complex conjugates. Hence, their product $[x-(1+i)][x-(1-i)]=|x-(1+i)|^2$ is always nonnegative.
Therefore, $f(x)$ is always positive when $x$ is real, and so the range of $f(x)$ is the set of all positive real numbers. In interval notation, this is represented as $(0,\infty)$. Hence, $a+b=\boxed{\infty}$.
To find the range of the function $f(x) = \frac{2}{2+4x^2-4x}$, we can start by determining the domain of the function, which is the set of all possible values for $x$.
In this case, the function is defined for all real numbers except the values that make the denominator equal to zero. So, let's solve the equation $2+4x^2-4x=0$:
$4x^2 - 4x + 2 = 0$
We can use the quadratic formula to solve this equation: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, with $a=4$, $b=-4$, and $c=2$.
Plugging these values into the formula, we get:
$x = \frac{-(-4) \pm \sqrt{(-4)^2-4(4)(2)}}{2(4)}$
Simplifying further:
$x = \frac{4 \pm \sqrt{16-32}}{8}$
$x = \frac{4 \pm \sqrt{-16}}{8}$
Since the square root of a negative number is not a real number, there are no real solutions to this equation. Therefore, the denominator $2+4x^2-4x$ is never equal to zero, and the function is defined for all real numbers.
Next, let's examine the behavior of the function as $x$ gets infinitely large in both positive and negative directions.
As $x$ approaches infinity, both $4x^2$ and $-4x$ become infinitely large, but the term $2$ remains constant. Therefore, $2+4x^2-4x$ approaches positive infinity. As a result, $f(x)=\frac{2}{2+4x^2-4x}$ approaches zero.
Similarly, as $x$ approaches negative infinity, $2+4x^2-4x$ also approaches positive infinity, and $f(x)$ approaches zero.
Since the function $f(x)$ never equals zero, the range of $f(x)$ is all real numbers except zero. In interval notation, we can write this as $(-\infty,0)\cup(0,\infty)$. The closed interval form of this range is $(-\infty,0] \cup [0,\infty)$.
Finally, $a$ is $-\infty$ and $b$ is $0$. Therefore, $a+b = -\infty + 0 = -\infty$.
To find the range of the function $f(x)$, we need to determine the values that $f(x)$ can take as $x$ varies over its domain.
First, let's find the domain of the function. Since $f(x)$ involves fractions, we need to ensure that the denominator is not equal to zero.
Setting the denominator $2+4x^2-4x$ equal to zero and solving for $x$, we get:
$$2+4x^2-4x = 0$$
Rearranging this quadratic equation, we have:
$$4x^2 - 4x + 2 = 0$$
Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with $a = 4$, $b = -4$, and $c = 2$, we can calculate the values of $x$:
$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(2)}}{2(4)}$$
$$x = \frac{4 \pm \sqrt{16-32}}{8}$$
$$x = \frac{4 \pm \sqrt{-16}}{8}$$
Since the discriminant is negative ($-16 < 0$), the quadratic equation has no real solutions. Hence, the denominator $2+4x^2-4x$ is never equal to zero, and the function $f(x)$ is defined for all real numbers.
Therefore, there are no restrictions on the values that $f(x)$ can take, and the range of the function is the set of all real numbers.
Hence, the interval $(a, b]$ can be written as $(-\infty, \infty]$, and $a+b = -\infty + \infty$ is not defined.