# Thread: Precise Definition of a: f(x)=x^2, L=49, c=-7, epsilon=0.4

1. ## Precise Definition of a: f(x)=x^2, L=49, c=-7, epsilon=0.4

For the given value of f(x) and values of L, c, and epsilon>0, find the largest open interval about c on which the inequality absolute value of f(x)-L<epsilon holds.
f(x)=x^2 L=49 c=-7 epsilon=0.4

here's the work I did (sorry this is my first post and I couldn't figure out how to do special characters or absolute value symbols)
x^2 -49<.4
48.6<x^2<49.4
6.9713<x<7.0285
So I thought my interval would be (6.9713,7.0285) but the answer was supposed to be (-7.0285,6.9713)
Where did I go wrong?

2. Originally Posted by Anastasia3rd
For the given value of f(x) and values of L, c, and epsilon>0, find the largest open interval about c on which the inequality absolute value of f(x)-L<epsilon holds.
f(x)=x^2 L=49 c=-7 epsilon=0.4

here's the work I did (sorry this is my first post and I couldn't figure out how to do special characters or absolute value symbols)
x^2 -49<.4
48.6<x^2<49.4
6.9713<x<7.0285
So I thought my interval would be (6.9713,7.0285) but the answer was supposed to be (-7.0285,6.9713)
Where did I go wrong?
Perhaps spend some more time on the Absolute Values and the Squares.

Is x < 0, potentially?

3. Originally Posted by Anastasia3rd
For the given value of f(x) and values of L, c, and epsilon>0, find the largest open interval about c on which the inequality absolute value of f(x)-L<epsilon holds.
f(x)=x^2 L=49 c=-7 epsilon=0.4

here's the work I did (sorry this is my first post and I couldn't figure out how to do special characters or absolute value symbols)
x^2 -49<.4
48.6<x^2<49.4
6.9713<x<7.0285
So I thought my interval would be (6.9713,7.0285) but the answer was supposed to be (-7.0285,6.9713)
Where did I go wrong?
No wonder you are confused. Either you misread the answer or the book has a typographical error.

$|x^2 - (-\ 7)^2| < 0.4 \implies |x^2 - 49| = 0.4 \implies - \ 0.4 < x^2 - 49 < 0.4 \implies$

$48.6 < x^2 < 49.4 \implies 6.9713 < x < 7.0285 \text { or } -\ 7.0285 < x < - \ 6.9713.$

But we are looking for a neighborhood close to MINUS 7. So the first neighborhood is irrelevant.

Thus we get $-\ 7.0285 < x < - \ 6.9713$ as the answer.

It is quite clear that $-\ 7.0285 < x < 6.9713$ is an error.

x = 0 lies in that interval and

$|0^2 - 49| = 49 \not < 0.4.$

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