Homework: DC circuit help
By applying the superposition theorem to your circuit I have noticed that the current you are looking for can only be dependent from voltage source E, because when you are looking at another source, E is shorted, so your current gets 0A.  So now it depends on E alone (all other sources shorted or open). 


Ig is open
Ig4 is open
E4 is shorted


So R1 gets in serial to R4 and R2 and R3 get just serial, too. Finally the serial resistance groups are in parallel. 

R14=2 Ω + R4
R23=8 Ω
=&gt; R14|23 = (16 Ω + 8R4)/(10 Ω + R4)
I=-(E / R14|23)
 =-(50 Ω + 5R4)/(16 Ω + 8R4) A
=&gt; R4=(-50 ΩA - 16I Ω )/(8I + 5 A)

When I=-3 A:
R4=2/19 Ω ≈ 0.1 Ω
When I=-1 A:
R4=34/3 Ω ≈ 11.3 Ω


As R4 is effectively rising with I, it must be that R4∈[0.1 Ω; 11.3 Ω]. 

Hope this helped a bit, though late....