## anonymous 4 years ago for f(x) = x^2/2, find the largest possible delta such that if 0 < |x+4| < delta, then |(x^2/2) - 8| < 1

1. anonymous

I have figured out to factor $x^{2}$ - 8 to get 1/2 (x-4)(x +4)...I hope that's correct...to make a correlation with delta...

2. anonymous

x^2/2 - 8, that is

3. JamesJ

First let's analyze |(x^2/2) - 8| < 1 This is equivalent to | x^2 - 16 | < 2 -2 < x^2 - 16 < 2 14 < x^2 < 18 For x < 0, -sqrt(14) > x > -sqrt(18) 4 - sqrt(18) < x + 4 < 4 - sqrt(14) and this is implied by | x + 4 | < |4 - sqrt(18)| because | 4 - sqrt(18) | < | 4 - sqrt(14) | Hence, given $\delta = \sqrt{18} - 4$ $| x + 4 | < \delta \implies \left| \frac{x^2}{2} - 8 \right| < 1$

4. anonymous

thanks!

5. JamesJ

You should draw a graph of y = x^2/2 over the interval |x + 4| < delta and see exactly how this delta works.