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.

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

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

But we are looking for a neighborhood close to

**MINUS 7. **So the first neighborhood is irrelevant.

Thus we get \(\displaystyle -\ 7.0285 < x < - \ 6.9713\) as the answer.

It is quite clear that \(\displaystyle -\ 7.0285 < x < 6.9713\) is an error.

x = 0 lies in that interval and

\(\displaystyle |0^2 - 49| = 49 \not < 0.4.\)