Proving Uniform Continuity of f(x): Let x in [Infinity, 0)

In summary, the problem asks to prove that the function f(x) = (x^2)/(1+x) for all x in [0, infinity) is uniformly continuous. The attempt at a solution involved using the definition of a uniform continuous function, which states that for any given epsilon > 0, there exists a delta > 0 such that for any u and v in the domain, if |u-v| < delta, then |f(u)-f(v)| < epsilon. To simplify the problem, long division was used to obtain the expression (x-1) - 1/(x+1). The crucial point to prove uniform continuity is to show that the delta only depends on epsilon, not on any specific points
  • #1
minyo
2
0

Homework Statement



let f(x)= (x^2)/(1+x) for all x in [ifinity, 0) proof that f(x) is uniformly continuous. can anyone help me with this problem

Homework Equations



using the definition of a uniform continuous function

The Attempt at a Solution



i did long division to simplify the problem and got (x-1) - 1/(x+1)
 
Physics news on Phys.org
  • #2
What is the definition of "uniformly continuous" and how have you tried to use it?
 
  • #3
Find a delta that doesn't depend on any specific point in the domain of f.
 
  • #4
for u and v in D,

|f(u)-f(v)|<epsilon if |u-v|<delta
 
  • #5
Do you mean the domain to be [0, infinity) or (0, infinity)? Since "infinity" isn't a real number, "[infinity, 0)" doesn't make sense. I suspect it was supposed to be [0, infinity).

"for u and v in D,

|f(u)-f(v)|<epsilon if |u-v|<delta"

is much too "shorthand" to be useful here. f is uniformly continuous in D if f(x) is defined for all x in d and given epsilon> 0, there exist delta> 0 such that if |u-v|< delta, then |f(u)- f(v)|< epsilon.

The crucial point is, as JG89 said, the delta depends only on epsilon, not u or v.

|f(u)- f(v)|= |u^2/(1+u)- v^2/(1+v)|= |(u^2(1+v)- v^2(1+u))/(1+ u+ v+ uv)|
= |u^2- v^2+ uv(u- v))/(1+ u+ v+ uv)|= |(u+v- uv)/(1+u+v+uv)||u-v|.

If you can find an upper bound for that first fraction, you are home free.
 

1. What is uniform continuity?

Uniform continuity is a property of a function that states that for any given value of x, the function will not vary significantly when x is changed by a small amount. In other words, the function maintains a relatively constant rate of change over a given interval.

2. How is uniform continuity different from continuity?

Uniform continuity is a stronger version of continuity. While continuity requires that a function be continuous at every point in its domain, uniform continuity also requires that the function's rate of change does not vary significantly over a given interval.

3. How can I prove that a function is uniformly continuous?

To prove that a function is uniformly continuous, you can use the epsilon-delta definition. This involves showing that for any given value of epsilon (representing a small change in the function's output), there exists a corresponding value of delta (representing a small change in the function's input) that ensures the function's output will not vary by more than epsilon.

4. What is the importance of proving uniform continuity?

Proving uniform continuity is important because it guarantees that a function will behave in a predictable and consistent manner over a given interval. This allows us to make accurate predictions and calculations using the function.

5. Can a function be uniformly continuous on an open interval?

No, a function must be defined and continuous on a closed interval in order to be uniformly continuous. This is because uniform continuity requires the function to have a defined limit at the endpoints of the interval.

Similar threads

  • Calculus and Beyond Homework Help
Replies
26
Views
891
  • Calculus and Beyond Homework Help
Replies
4
Views
301
  • Calculus and Beyond Homework Help
Replies
3
Views
976
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
898
  • Calculus and Beyond Homework Help
Replies
13
Views
11K
Back
Top