How to determine the x values where a function is continuous

In summary, the conversation discusses the continuity of a function, specifically looking at the x values where it is continuous and the types of discontinuities that may occur. The function in question is not defined at x = 1 and x = -1, but is continuous at all other points. To study the behavior of the function near these singularities, limits are calculated and it is found that the left and right limits are different. The function is therefore technically not continuous at these points, but can be written using notation to show that the limits approach infinity from opposite directions. The conversation also discusses the possibility of other types of discontinuities and interval notation.
  • #1
TheRedDevil18
408
1
How would I find the x values for which a function is continuous ?, and how to tell whether it is a removable discontinuity, a jump discontinuity, or an infinite discontinuity ?

Suppose the function is sqrt(9-x^2)/x^2-1
 
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  • #2
[itex]\frac{\sqrt{9-x^2}}{x^2-1}=\frac{\sqrt{9-x^2}}{(x-1)(x+1)}[/itex]. So your function is not defined on the points 1 and -1, but it is continuous on the other points as a composition of basic continuous functions. To study the behavior of the function near the singularities, you calculate the limits, you will find [tex]\lim_{x\rightarrow 1} \frac{\sqrt{9-x^2}}{x^2-1}=+ \infty[/tex] from the right and [itex]-\infty[/itex] from the left. The opposite thing happens for -1.
 
  • #3
ndjokovic said:
[itex]\frac{\sqrt{9-x^2}}{x^2-1}=\frac{\sqrt{9-x^2}}{(x-1)(x+1)}[/itex]. So your function is not defined on the points 1 and -1, but it is continuous on the other points as a composition of basic continuous functions. To study the behavior of the function near the singularities, you calculate the limits, you will find [tex]\lim_{x\rightarrow 1} \frac{\sqrt{9-x^2}}{x^2-1}=+ \infty[/tex] from the right and [itex]-\infty[/itex] from the left. The opposite thing happens for -1.

Wouldn't the numerator also have a restriction ?, the square root can't be negative. So would it be correct in saying that the graph is continuous on the intervals [-3,-1) U [3,1) ?
 
  • #4
f is continuous WHEREVER it is DEFINED (i.e, has a function value).
 
  • #5
TheRedDevil18 said:
Wouldn't the numerator also have a restriction ?, the square root can't be negative. So would it be correct in saying that the graph is continuous on the intervals [-3,-1) U [3,1) ?
Interval notation goes left to right, so this would be written as [-3, -1) U (1, 3].
 
  • #6
ndjokovic said:
[itex]\frac{\sqrt{9-x^2}}{x^2-1}=\frac{\sqrt{9-x^2}}{(x-1)(x+1)}[/itex]. So your function is not defined on the points 1 and -1, but it is continuous on the other points as a composition of basic continuous functions. To study the behavior of the function near the singularities, you calculate the limits, you will find [tex]\lim_{x\rightarrow 1} \frac{\sqrt{9-x^2}}{x^2-1}=+ \infty[/tex] from the right and [itex]-\infty[/itex] from the left. The opposite thing happens for -1.
Technically, ##\lim_{x\to 1} \frac{\sqrt{9-x^2}}{x^2-1} ## does not exist, since the left and right limits are different (which is what you said). The notation that is often used is
##\lim_{x \to 1^+} \frac{\sqrt{9-x^2}}{x^2-1} = \infty## and ##\lim_{x \to 1^-} \frac{\sqrt{9-x^2}}{x^2-1} = -\infty##.
 
  • #7
The function is perfectly continuous at (-1,1) as well.
 

1. How do I know if a function is continuous at a specific value of x?

The function is continuous at a specific value of x if the limit of the function at that value is equal to the value of the function at that point.

2. What is the difference between a removable and non-removable discontinuity?

A removable discontinuity occurs when there is a hole in the graph of the function at a specific value of x, but the limit of the function at that value exists. A non-removable discontinuity occurs when there is a jump or break in the graph of the function at a specific value of x, and the limit of the function at that value does not exist.

3. Can a function be continuous at one value of x and discontinuous at another?

Yes, it is possible for a function to be continuous at one value of x and discontinuous at another. This occurs when there is a removable discontinuity or a jump in the graph of the function at a specific value of x.

4. What is a piecewise function and how does it affect continuity?

A piecewise function is a function that is defined by different equations for different intervals of x. It can affect continuity if there are discontinuities at the points where the equations change, or if the functions are not defined at these points.

5. How can I determine continuity at the endpoints of a closed interval?

To determine continuity at the endpoints of a closed interval, you must evaluate the function at the endpoints and check if the limit of the function at those points exists and is equal to the value of the function at those points. If the function is defined on an open interval, then the limit of the function at the endpoint must also be checked to determine continuity.

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