Consider the function below. lim x -> 3 (2x+1)=7

What values of x guarantee that f(x)=2x+1 is within 0.04 units of 7?

If x is within _____ units of 3, then f(x) is within 0.04 units of 7.
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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.


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. 

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