text: # Defining a function and a delta limit problem
  
  1. Jun 2, 2009
  
  ### Jimmy84
  
  1. The problem statement, all variables and given/known data
  
  Im having a lot of dificulties evaluating this function, I really need some easy to understand explanation about how to evaluate it by the given values, I would appreciate any help.
  
  -Problem one
  
  given the function:
  
  U(x) =
  
  0 if x < 0
  1 if 0 ≤
  
  define the function by
  
  a.) U(x-1)
  b.) U(x) - 1
  c.) U(x) - U(x-1)
  
  The solutions are
  
  a.) U(x-1) =
  
  0 if x < 1
  1 if 1 ≤ x
  
  b.) U(x) - 1 =
  
  -1 if x < 0
  0 if 0 ≤ x
  
  c.) U(x) - U(x-1) =
  
  0 if x < 0
  1 if 0 ≤ x < 1
  0 if 1 ≤ x
  
  -Problem 2
  given f(x), a, L and epsilon, determine that delta > 0
  
  the given function is x^2 -2x +1 , a = 2 , L = 1 and epsilon = 0.4
  
  so I tried getting f(x1) and f(x2)
  
  f(x1) = 0.6 and f(x2) = 1.4
  
  but then I got stuck with the algebra when I had to get x1 and x2.
  
  (x1)^2 - 2(x1) + 1 = o.6
  
  and
  
  (x2)^2 - 2(x2) + 1 = 1.4
  
  how can I solve these problems?
  
  2. Jun 2, 2009
  
  ### Staff: Mentor
  
  For #1, have you sketched a graph of the function? The graph consists two horizontal lines. If you know what the graph of y = f(x) looks like, the graph of y = f(x -c) (where c is a positive number) is the translation to the right of y = f(x) by c units.
  
  The graph of y = f(x) - c, (again c > 0), is the translation down of the graph of y = f(x) by c units.
  
  For #2, the problem as you state it doesn't make much sense:
  From the rest of what you show, it appears that you are trying to establish that
  $$\lim_{x \rightarrow 2} x^2 - 2x + 1 = 1$$
  
  Draw a graph of y = x2 - 2x + 1 if you haven't already done so. You are given an epsilon of .4, so you are trying to find a value for delta, so that if |x - 2| < delta, then |f(x) - 1| < .4.
  
  The equations x12 - 2x1 + 1 = .6 and x22 - 2x2 + 1 = 1.4 are quadratic equations. Surely you have had practice in solving quadratic equations if you are starting in on delta-epsilon proofs of limits. After you have found x1 and x2, take delta to be the smaller of the differences |x1 - 2| and |x2 - 2|.
  
  3. Jun 2, 2009
  
  ### HallsofIvy
  
  Staff Emeritus
  In the first one, you have just replaced x by x-1 so in the conditions x< 0, 0< x, replace that x by x-1 also: x-1< 0 and 0< x-1. What does that simplify to?
  
  4. Jun 3, 2009
  
  ### Jimmy84
  
  Thanks a lot guys I already have the graphs for every problem. Im still having dificulties with problem 2.
  
  Im not sure about how to solve this:
  
  (x1)^2 - 2(x1) + 1 = 0.6
  
  and
  
  (x2)^2 - 2(x2) + 1 = 1.4
  
  I tried doing this but I dont think that it is right.
  
  (x1)^2 - 2(x1) + 1 = 0.6
  
  (x1)^2 - 2(x1) = -0.4
  
  (x1)^2 = 0.2
  
  x1 = square root of 0.2
  
  x1 = 0.447
  
  and I couldnt solve (x2)^2 - 2(x2) + 1 = 1.4
  
  5. Jun 3, 2009
  
  ### HallsofIvy
  
  Staff Emeritus
  It would help a lot to recognize that $x^2- 2x+ 1= (x- 1)^2$.
  
  6. Jun 3, 2009
  
  ### Staff: Mentor
  
  Or it would help if you reviewed how to solve quadratic equations.
  How did you get this? What happened to the -2x1 term and the -.4 term?
  If you're trying to solve the equation x12 -2x1 + 1 = .6, HallsOfIvy's suggestion of recognizing the perfect square trinomial on the left side is the quickest way. If you don't recognize that, you can always use the quadratic formula, after a bit of rearrangement.
  
  x12 - 2x1 +1 = 0.6
  ==> x12 - 2x1 + 0.4 = 0
  
  Now you can use the quadratic formula, with a = 1, b = -2, and c = 0.4.
explanation: The document contains a math question and a partial attempt at an answer. The user is asking for help in solving two specific problems related to functions and limits. The user has provided the problem statements and their attempts at solving them, but is stuck on solving quadratic equations.
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distilabel_metadata.role: system
distilabel_metadata.content: # Defining a function and a delta limit problem
  
  1. Jun 2, 2009
  
  ### Jimmy84
  
  1. The problem statement, all variables and given/known data
  
  Im having a lot of dificulties evaluating this function, I really need some easy to understand explanation about how to evaluate it by the given values, I would appreciate any help.
  
  -Problem one
  
  given the function:
  
  U(x) =
  
  0 if x < 0
  1 if 0 ≤
  
  define the function by
  
  a.) U(x-1)
  b.) U(x) - 1
  c.) U(x) - U(x-1)
  
  The solutions are
  
  a.) U(x-1) =
  
  0 if x < 1
  1 if 1 ≤ x
  
  b.) U(x) - 1 =
  
  -1 if x < 0
  0 if 0 ≤ x
  
  c.) U(x) - U(x-1) =
  
  0 if x < 0
  1 if 0 ≤ x < 1
  0 if 1 ≤ x
  
  -Problem 2
  given f(x), a, L and epsilon, determine that delta > 0
  
  the given function is x^2 -2x +1 , a = 2 , L = 1 and epsilon = 0.4
  
  so I tried getting f(x1) and f(x2)
  
  f(x1) = 0.6 and f(x2) = 1.4
  
  but then I got stuck with the algebra when I had to get x1 and x2.
  
  (x1)^2 - 2(x1) + 1 = o.6
  
  and
  
  (x2)^2 - 2(x2) + 1 = 1.4
  
  how can I solve these problems?
  
  2. Jun 2, 2009
  
  ### Staff: Mentor
  
  For #1, have you sketched a graph of the function? The graph consists two horizontal lines. If you know what the graph of y = f(x) looks like, the graph of y = f(x -c) (where c is a positive number) is the translation to the right of y = f(x) by c units.
  
  The graph of y = f(x) - c, (again c > 0), is the translation down of the graph of y = f(x) by c units.
  
  For #2, the problem as you state it doesn't make much sense:
  From the rest of what you show, it appears that you are trying to establish that
  $$\lim_{x \rightarrow 2} x^2 - 2x + 1 = 1$$
  
  Draw a graph of y = x2 - 2x + 1 if you haven't already done so. You are given an epsilon of .4, so you are trying to find a value for delta, so that if |x - 2| < delta, then |f(x) - 1| < .4.
  
  The equations x12 - 2x1 + 1 = .6 and x22 - 2x2 + 1 = 1.4 are quadratic equations. Surely you have had practice in solving quadratic equations if you are starting in on delta-epsilon proofs of limits. After you have found x1 and x2, take delta to be the smaller of the differences |x1 - 2| and |x2 - 2|.
  
  3. Jun 2, 2009
  
  ### HallsofIvy
  
  Staff Emeritus
  In the first one, you have just replaced x by x-1 so in the conditions x< 0, 0< x, replace that x by x-1 also: x-1< 0 and 0< x-1. What does that simplify to?
  
  4. Jun 3, 2009
  
  ### Jimmy84
  
  Thanks a lot guys I already have the graphs for every problem. Im still having dificulties with problem 2.
  
  Im not sure about how to solve this:
  
  (x1)^2 - 2(x1) + 1 = 0.6
  
  and
  
  (x2)^2 - 2(x2) + 1 = 1.4
  
  I tried doing this but I dont think that it is right.
  
  (x1)^2 - 2(x1) + 1 = 0.6
  
  (x1)^2 - 2(x1) = -0.4
  
  (x1)^2 = 0.2
  
  x1 = square root of 0.2
  
  x1 = 0.447
  
  and I couldnt solve (x2)^2 - 2(x2) + 1 = 1.4
  
  5. Jun 3, 2009
  
  ### HallsofIvy
  
  Staff Emeritus
  It would help a lot to recognize that $x^2- 2x+ 1= (x- 1)^2$.
  
  6. Jun 3, 2009
  
  ### Staff: Mentor
  
  Or it would help if you reviewed how to solve quadratic equations.
  How did you get this? What happened to the -2x1 term and the -.4 term?
  If you're trying to solve the equation x12 -2x1 + 1 = .6, HallsOfIvy's suggestion of recognizing the perfect square trinomial on the left side is the quickest way. If you don't recognize that, you can always use the quadratic formula, after a bit of rearrangement.
  
  x12 - 2x1 +1 = 0.6
  ==> x12 - 2x1 + 0.4 = 0
  
  Now you can use the quadratic formula, with a = 1, b = -2, and c = 0.4.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a partial attempt at an answer. The user is asking for help in solving two specific problems related to functions and limits. The user has provided the problem statements and their attempts at solving them, but is stuck on solving quadratic equations.
  
  Label: question-and-answer