anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
so if you were to give an exact number for delta it would be 2.4?
anonymous
  • anonymous
no 2.3 sorry

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jim_thompson5910
  • jim_thompson5910
what 0 < |x – 2| < δ means is that x is 2 less than δ units away from 2
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
which means that the max δ can be is 0.4
anonymous
  • anonymous
ooh ok thanks so much
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
it said this was wrong
anonymous
  • anonymous
no its right thanks you
jim_thompson5910
  • jim_thompson5910
hmm i was gonna say lol
jim_thompson5910
  • jim_thompson5910
yw

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