Question

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

Answer 1

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...

Answer 2

x^2/2 - 8, that is

Answer 3

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
\[ \delta = \sqrt{18} - 4 \]
\[ | x + 4 | < \delta \implies \left| \frac{x^2}{2} - 8 \right| < 1 \]

Answer 4

thanks!

Answer 5

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