Let x=3±h where h is small, then f(3±h)=2(3±h)+1=7±2h, and this must equal 7±0.04.
So we remove 7 from each side, leaving 2h=±0.04, and h=±0.02.
If x is within 0.02 units of 3, then f(x) is within 0.04 units of 7. Strictly, 2.98≤x≤3.02 when 6.96≤f(x)≤7.04.
CHECK
f(3.02)=2×3.02+1=7.04 and f(2.98)=2×2.98+1=6.96. Both 6.96 and 7.04 are within 0.04 units of 7, including the actual values 6.96 and 7.04.