Problem 869

For the given function f(x) and values of L, c, and ε > 0 find the largest open interval about c on which the inequality |f(x) – L| < ε holds. Then determine the largest values for δ > 0 such that 0 < |x – c| < δ → |f(x) – L| < ε.

f(x) = 2x + 3, L = 9, c = 3, ε = 0.04

 

Solution:-

 

Solve |f(x) – L| ε to find the largest interval containing c on which the inequality holds.

|(2x+3)-9|<0.04 → -0.04 <[(2x+3)-9]<0.04

Combine the constants.

-0.04 < 2x -6 <0.04

Simplify

5.96 < 2x < 6.04

Complete the solution of the inequality by dividing the three expressions by 2.

2.98 < x < 3.02

Since the interval 2.98 < x < 3.02 is centered on c = 3, δ  is the distance from 3 to either endpoint of the interval.

Thus , the value of δ is 0.02.

 

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