The range of the function $f(x) = \frac{2}{2+4x^2-4x}$ can be written as an interval $(a,b]$. What is $a+b$?

Answer 1

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, $a+b=\boxed{\infty}$.