Answer:
D
Step-by-step explanation:
We have the piecewise function:

And we want to find x such that f(x)=1.
So, let's substitute 1 for f(x):

This has two equations. So, we can separate them into two separate cases. Namely:

Let's solve for x in each case.
Case I:
We have:

Add 1 to both sides:

Let's cancel out the fraction by multiplying both sides by -2. So:

The right side cancels:

Flip:

So, x is -4.
Case II:
We have:

Flip:

This is the solution for our second case.
So, we have:

Now, can check to see if we have to to remove solution(s) that don't work.
Note that x=-4 is the solution to our first equation.
The first equation is defined only if x is less than -2.
-4 <em>is </em>less than -2. So, x=-4 is indeed a solution.
x=1 is the solution to our second equation.
The second equation is defined only if x is greater than or equal to -2.
1 <em>is</em> greater than or equal to -2. So, x=1 is <em>also </em>a solution.
Therefore, our two solutions are:

Out of our answer choices, we can pick D.
And we're done!