Think positive

First, we observe that all 3 components of f(x) -- $2, |x|, x^2$ -- are symmetric around x =0. So f(-x) = f(x).

Therefore, we only need to look at the case where x > 0 (at 0 itself, f(x) is positive as the two components involving x vanish, leaving f(0) = 2.) For x > 0, |x| = x.

We observe that f'(x) = 1 - 2x and f''(x) = -2. so the function starts positive and once it becomes negative, it never becomes positive again.

For positive x, factorizing yields (2-x)(x-1) = 0 if we're trying to find zeroes. So x = -1 (impossible) or x = 2. The function turns negative as it crosses x = 2, and hence when it crosses x = -2 as well.

So the range where it is positive is -2 < x < 2

$f(x) = 2 + |x| - x^2 =\begin{cases} 2 + x - x^2 & \quad\quad x \geq 0\\ 2 - x - x^2 & \quad\quad x<0\\ \end{cases}$

We have two cases to analyze.

The first case: $2+x-x^2 > 0$

$(2-x)(1+x) > 0$

Draw out a number line or the graph of this quadratic function, and you should get the range of:

$-1<x<2$

However, this function is only defined for $x \geq 0$

Therefore, the range of values of $x$ becomes $0 \leq x < 2$

The second case: $2-x-x^2 > 0$

$(2+x)(1-x) > 0$

Draw out a number line or the graph of this quadratic function, and you should get the range of:

$-2<x<1$

However, this function is only defined for $x < 0$

Therefore, the range of values of $x$ becomes $-2 < x < 0$

Combine the two ranges of values of $x$ , and you will get the final answer:

$\boxed{-2 < x < 2}$