Question

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

f(x) = 3x + 5, L=14, c-3, ∊ = 0.12
The largest open interval about c on which the inequality |f(x)-L|< ∊ holds is ______ The largest value of δ > 0 such that 0 < |x-c|< δ → |f(x)-L| < ∊ is ______

Answer 1

The largest open interval about c on which the inequality |f(x)-L| < ∊ holds is (2.9867, 3.0133). The largest value of δ > 0 such that 0 < |x-c| < δ → |f(x)-L| < ∊ is 0.0133.

Step-by-step explanation:

To find the largest open interval about c on which the inequality |f(x)-L| < ∊ holds, we need to find the range of values for x for which |3x + 5 - 14| < 0.12 holds. Solving this inequality, we have |3x - 9| < 0.12. This inequality is satisfied when -0.04 < 3x - 9 < 0.04, which gives -0.04 + 9 < 3x < 0.04 + 9. Simplifying further, we get 8.96 < 3x < 9.04. Dividing by 3, we have 2.9867 < x < 3.0133. Therefore, the largest open interval about c is (2.9867, 3.0133).

To determine the largest value for δ > 0 such that 0 < |x-c| < δ → |f(x)-L| < ∊, we need to find the range of values for x close to c that satisfy |3x + 5 - 14| < 0.12. Given that c = 3, we can rewrite the inequality as |3x - 9| < 0.12. By definition, |x - c| represents the distance between x and c. Therefore, we want to find the largest value of δ such that if |x - c| < δ, then |3x - 9| < 0.12. It can be observed that as x approaches 3, the value of |3x - 9| also approaches 0. Therefore, we can choose δ to be any positive number less than the minimum distance between x and 3, which is the distance from c to the endpoints of the largest open interval about c. From the previous calculation, this distance is 3 - 2.9867 = 0.0133. Thus, the largest value of δ is 0.0133.