Solution Found!
Answer: Each of Exercises 15–30 gives a function ƒ(x) and
Chapter 2, Problem 17E(choose chapter or problem)
Each of Exercises 15 - 30 gives a function \(f(x)\) and numbers \(L\), \(c\), and \(\epsilon>0\). In each case, find an open interval point about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds.
\(f(x)=\sqrt{x+1}\), \(L = 1\), \(c = 0\), \(\epsilon=0.1\)
Equation Transcription:
Text Transcription:
f(x)
L
c
epsilon > 0
|f(x) - L| < epsilon
delta > 0
x
0 < |x-c| <
|f(x) - L| < epsilon
f(x) = x + 1, L = 1, c = 0, epsilon = 0.1
Questions & Answers
QUESTION:
Each of Exercises 15 - 30 gives a function \(f(x)\) and numbers \(L\), \(c\), and \(\epsilon>0\). In each case, find an open interval point about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds.
\(f(x)=\sqrt{x+1}\), \(L = 1\), \(c = 0\), \(\epsilon=0.1\)
Equation Transcription:
Text Transcription:
f(x)
L
c
epsilon > 0
|f(x) - L| < epsilon
delta > 0
x
0 < |x-c| <
|f(x) - L| < epsilon
f(x) = x + 1, L = 1, c = 0, epsilon = 0.1
ANSWER:Solution:-
Step 1 of 3
Given that
We have to find an open interval about on which the inequality holds, and then we have to give a value for
such that for all x satisfying
the inequality
holds.