The discussion revolves around determining the largest number that can fill in the blank in the statement regarding the proximity of x to 3 and the behavior of the function f(x) near that point. The subject area includes concepts of limits and continuity in calculus.
The conversation is ongoing, with participants exploring different interpretations of the function and its properties. Some guidance has been offered regarding the nature of the discontinuity and its implications for the problem, but no consensus has been reached on the correct approach or answer.
There is a noted lack of clarity regarding the function f(x) and its behavior at the point of discontinuity, which is central to resolving the problem. The original poster's mention of specific values (1.99 and 2.01) indicates an attempt to relate these to the function's output, but the discussion highlights the need for further clarification on the function itself.
This question doesn't make sense unless you are given a specific function f! It looks to me like f is some continuous function such that f(3)= 2. But what goes in the blank depends upon exactly what f is. Surely you realized that?MellowOne said:Homework Statement
For the statement "If x is within _______ units of 3 (but not equal to 3), then f(x) is within 0.01 unit of 2," write the largest number that can go in the blank
Homework Equations
Nonr
The Attempt at a Solution
It's a multi-step problem, but the two numbers I ended up with are 1.99 or 2.01. I'm not sure which one fits into this statement because I'm not quite sure what it's asking.
The exclamation point after f was punctuation in the sentence, just like this one!MellowOne said:Sorry, maybe I should have put the function that I was given. I'm not sure what F! means, but here's the function I'm given. F(x) = (x^3 - 7x^2 + 17x - 15)/( x - 3) It is a continuous graph, but at x = 3 there's a point of discontinuity which is (3,2).
You mean that F has a removable discontinuity at x= 3. That's a discontinuity because 33- 7(32+ 17(3)- 15= 27-63+ 51- 15= 78- 78= 0 as well as 3- 3= 0: both numerator and denominator are 0 at x= 3. Because x= 3 makes the numerator 0, we know that x-3 is a factor. Knowing that it is easy to see that x3- 7x2+ 17x- 15= (x-3)(x2- 4x+ 5) so for x NOT equal to 3, this is just F(x)= x2- 4x+ 5= (x-4)(x-1). The discontinuity is "removable" because that has limit 3 at x= 2.MellowOne said:Sorry, maybe I should have put the function that I was given. I'm not sure what F! means, but here's the function I'm given. F(x) = (x^3 - 7x^2 + 17x - 15)/( x - 3) It is a continuous graph, but at x = 3 there's a point of discontinuity which is (3,2).