Filling in the blank to a statement

  • Thread starter MellowOne
  • Start date
In summary: Well, we can solve that for x. Multiply both sides by -1 to get -2.01\le -x2+ 4x- 5\le -1.99. Add 5 to all sides: 2.99\le -x2+ 4x\le 3.01. Now, complete the square in x: -x2+ 4x+ 4= -(x-2)2+ 4. So we have -x2+ 4x+ 4\le 3.01, -(x-2)2\le 3.01- 4= -0.99. Because the square of a number is
  • #1
MellowOne
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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.
 
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  • #2
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.
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?
 
  • #3
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).
 
  • #4
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).
The exclamation point after f was punctuation in the sentence, just like this one!
The graph can't be continuous AND have a point of discontinuity. Without looking at the graph, I'm guessing that there is a "hole" at (3, 2). I'm also guessing that if you divided the numerator polynomial (x^3 + ...) by (x - 3) there wouldn't be a remainder. That might be a hint.
 
  • #5
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.

Now, if "f(x) is within 0.01 unit of 2", that is, if [itex]1.99\le x^2- 4x+ 5\le 2.01[/itex] what must x be?
 

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