# Coming up with an epsilon value

• Aug 5th 2007, 11:42 AM
spiritualfields
Coming up with an epsilon value
I'm having a little problem with this. Here is the problem statement:
Quote:

Use the graph of the function f given in figure 3.21 to estimate a number epsilon such that: If 0 < |x - 1| < 0.5, then |f(x) - 1.5| < e. Find e.
Looking at the graph of the function, f(.5) is approx: .75, and f(1.5) is approx 1.75. So, within the desired interval the range of f(x) varies from .75 to 1.75. The correct answer is given as an epsilon = .75.
Quote:

But if |f(x) - 1.5| < .75, then -.75 < f(x) - 1.5 < .75, and this would give a range of f(x) as (.75, 2.25), and 2.25 is well out of the range of the interval.
So I don't understand this answer at all.
• Aug 5th 2007, 05:48 PM
spiritualfields
Well, 9 views (as of now) isn't exactly a quorum, but I'm not surprised at receiving no reply as of yet. With the values of (x,f(x)) that I got (.5, .75), and (1.5, 1.75), there's just no way epsilon can be .75 if the value that f(x) is being compared with is 1.5. For these values, this makes more sense:
Quote:

|f(x) - 1.25| < .50
With an epsilon of .50, and with a value of 1.25 instead of 1.50, this becomes:
Quote:

-.50 < f(x) - 1.25 < .50, so .75 < f(x) < 1.75, which would agree with the values of f(x) at .5 and at 1.5 (which is the interval of interest, as given in the problem statement).
So for now I'm just chalking it up to a misprint in the book or that I'm in serious need of eyeglasses (which I'm not).
• Aug 5th 2007, 07:39 PM
CaptainBlack
Quote:

I'm having a little problem with this. Here is the problem statement:
Quote:

Use the graph of the function f given in figure 3.21 to estimate a number epsilon such that: If 0 < |x - 1| < 0.5, then |f(x) - 1.5| < e. Find e.
Looking at the graph of the function, f(.5) is approx: .75, and f(1.5) is approx 1.75. So, within the desired interval the range of f(x) varies from .75 to 1.75. The correct answer is given as an epsilon = .75.
Quote:

But if |f(x) - 1.5| < .75, then -.75 < f(x) - 1.5 < .75, and this would give a range of f(x) as (.75, 2.25), and 2.25 is well out of the range of the interval.
So I don't understand this answer at all.
It runs the other way around, if 0 < |x - 1| < 0.5 what is the largest value for |f(x) - 1.5|? Well from what you say the largest value of this occurs when x=0.5, when |f(x) - 1.5|=0.75, so for all x in the range 0.5 to 1.7:

|f(x) - 1.5|<=0.75

so 0.75 will do for epsilon.

RonL
• Aug 5th 2007, 08:57 PM
spiritualfields
Quote:

It runs the other way around, if 0 < |x - 1| < 0.5 what is the largest value for |f(x) - 1.5|? Well from what you say the largest value of this occurs when x=0.5, when |f(x) - 1.5|=0.75, so for all x in the range 0.5 to 1.7:

|f(x) - 1.5|<=0.75

so 0.75 will do for epsilon.
Thanks for clearing that up. I see where I was wrong now. What I had been trying to do was get an epsilon that would be equidistant to 1.5 from the left and right side of it, and still be in the interval of (.5, 1.5). If I give up on that idea and just compare f(.5) to 1.5, then I get |.75 - 1.5| = .75. and if I compare f(1.5) to 1.5, then I get |1.75 - 1.5| = .25. So out of those two numbers, only .75 will work throughout the interval and still maintain the requirement that |f(x) - 1.5| < e.