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.