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.
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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 if you were to give an exact number for delta it would be 2.4?
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what 0 < |x – 2| < δ means is that x is 2 less than δ units away from 2
the furthest you can get from 2 is either x = 1.6 or x = 2.4
2.4 - 2 = 0.4
1.6 - 2 = -0.4
so this max distance you can get from x = 2 is 0.4 units
which means that the max δ can be is 0.4
ooh ok thanks so much
here's what is going on
x is getting closer and closer to 2
as that is happening, f(x) is getting closer and closer to 3
if you were to specify that f(x) can be no further away from f(x) = 3 than 0.4 units, you are saying this: |f(x) – 3| < 0.4
if you plug in f(x) = x+1 and solve for x, you get 1.6 < x < 2.4
that tells you that x must be between 1.6 and 2.4 and this is the boundary set up by the fact that |f(x) – 3| < 0.4
the value x is approaching is at the midpoint of the interval, so that is why the max distance you can get from this midpoint is the interval width cut in half (or the distance from the center x = 2 to x = 2.4)
so that explains why the max distance is δ = 0.4
hopefully that didn't make things murkier lol