Defining a function and a delta limit problem

[Jimmy84]

Homework Statement

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?


thanks a lot in advance.

 

[Mark44]

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:

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

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|.

 

[HallsofIvy]

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?

 

[Jimmy84]

Thanks a lot guys I already have the graphs for every problem. I am 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 don't 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 couldn't solve (x2)^2 - 2(x2) + 1 = 1.4

 

[HallsofIvy]

It would help a lot to recognize that $x^2- 2x+ 1= (x- 1)^2$.

 

[Mark44]

Or it would help if you reviewed how to solve quadratic equations.

(x1)^2 - 2(x1) + 1 = 0.6
(x1)^2 - 2(x1) = -0.4
(x1)^2 = 0.2

How did you get this? What happened to the -2x₁ term and the -.4 term?

x1 = square root of 0.2

x1 = 0.447

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.