The graph of f (x) = x + 1 is shown in the figure. Find the largest δ such that if 0 < |x - 2| < δ then |f (x) - 3| < 0.4.

SO what you need to do is:

Start with |f (x) - 3| < 0.4

and plug in f (x) = x+1

to get |f (x) - 3| < 0.4

|x+1 - 3| < 0.4

|x - 2| < 0.4

-0.4 < x - 2 < 0.4

-0.4+2 < x < 0.4+2

1.6 < x < 2.4

So delta would be 2.3

Hope this is what you were looking for