Let x=3±h where h is small.

f(3±h)=(3±h)²=9±6h+h²=9±0.2, so ±6h≈0.2 if we ignore h² as being insignificantly small.

So h=0.2/6=0.03 approx.

**If x is within 0.03 units of 3, then f(x) is within 0.2 units of 9.**

Also, **if x is within 0.033 units of 3, then f(x) is within 0.2 units of 9. **But it is not true if we add another 3, making h=±0.0333 because 3.0333²>9.2 even though 2.9667²>8.8 (that is, less than 0.2 difference from 9). The guaranteed value is 0.03 or 0.033. It’s always a good idea to check how many decimal places are safe. In this case, a maximum of 3 decimal places guarantees the correct variation for x.