Proof that a function is continuous on its domain

In summary: Thanks for the help!In summary, the homework statement is about proving that a function is continuous on its domain. The delta-epsilon definition of continuity is used to reduce the equation |f(x) - f(x_0)| to a simpler equation. The attempt at a solution is to factor x^2+x-2, but the question is where to go from there. Another way to prove cos is continuous on the domain is to use the trig identity.
  • #1
Whistlekins
21
0

Homework Statement



We have [itex]f(x) = \frac{x^{2}+x-2}{x-1}+cos(x) , x\in\mathbb{R}\setminus \{1\}[/itex] and wish to prove that it is continuous on its domain.

Homework Equations



The delta-epsilon definition of the continuity of a function.

The Attempt at a Solution



I've managed to reduce [itex]|f(x) - f(x_0)| [/itex]to[itex] |x-x_0| + |cos(x) - cos(x_0)| < \delta + |cos(x) - cos(x_0)|[/itex]
I'm not too sure where to go from there or even if I'm on the right track. Any insight would be greatly appreciated.
 
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  • #2
Whistlekins said:

Homework Statement



We have [itex]f(x) = \frac{x^{2}+x-2}{x-1}+cos(x) , x\in\mathbb{R}\setminus \{1\}[/itex] and wish to prove that it is continuous on its domain.

Homework Equations



The delta-epsilon definition of the continuity of a function.

The Attempt at a Solution



I've managed to reduce [itex]|f(x) - f(x_0)| [/itex]to[itex] |x-x_0| + |cos(x) - cos(x_0)| < \delta + |cos(x) - cos(x_0)|[/itex]
I'm not too sure where to go from there or even if I'm on the right track. Any insight would be greatly appreciated.

Factoring x^2+x-2 would be a great first step.
 
  • #3
Dick said:
Factoring x^2+x-2 would be a great first step.

I have, that's how I arrived at the reduced expression. The questions is where to go from there. I could always use the property that if [itex]f[/itex] and [itex]g[/itex] are continuous at a point [itex]x_0\in \mathbb{A}[/itex] then [itex]f+g[/itex] is continuous at [itex]x_0[/itex]. But I don't know how to prove that [itex]cos(x)[/itex] is continuous on the domain.
 
  • #4
Whistlekins said:
I have, that's how I arrived at the reduced expression. The questions is where to go from there. I could always use the property that if [itex]f[/itex] and [itex]g[/itex] are continuous at a point [itex]x_0\in \mathbb{A}[/itex] then [itex]f+g[/itex] is continuous at [itex]x_0[/itex]. But I don't know how to prove that [itex]cos(x)[/itex] is continuous on the domain.

Yes, that you did. One way to prove cos is continuous is to use the trig identity, cos(u)-cos(v)=(-2)sin((u+v)/2)*sin((u-v)/2).
 
  • #5
Dick said:
Yes, that you did. One way to prove cos is continuous is to use the trig identity, cos(u)-cos(v)=(-2)sin((u+v)/2)*sin((u-v)/2).

Ahh I overlooked that, thanks! I think I've got it now.
 

Related to Proof that a function is continuous on its domain

1. What is the definition of continuity for a function?

The definition of continuity for a function is that the function is considered continuous at a point if the limit of the function as x approaches that point exists and is equal to the value of the function at that point.

2. How do you prove that a function is continuous at a specific point?

To prove that a function is continuous at a specific point, you must show that the limit of the function as x approaches that point exists and is equal to the value of the function at that point. This can be done by using the definition of continuity and evaluating the limit algebraically or graphically.

3. What is the difference between pointwise continuity and uniform continuity?

Pointwise continuity refers to the continuity of a function at a specific point, while uniform continuity refers to the continuity of a function over an entire interval. A function can be pointwise continuous at every point in its domain, but not uniformly continuous over that same interval.

4. Can a function be discontinuous on its domain?

Yes, a function can be discontinuous on its domain. This means that there is at least one point in the domain where the function is not continuous. This could be due to a discontinuity in the function itself or a break in the domain of the function.

5. What are some common types of discontinuities in functions?

Some common types of discontinuities in functions include removable discontinuities, jump discontinuities, and infinite discontinuities. Removable discontinuities occur when there is a hole or gap in the graph of the function, while jump discontinuities occur when there is a sudden jump or break in the graph. Infinite discontinuities occur when the limit of the function at a certain point is either positive or negative infinity.

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