How to find continuous on the interval by graphing functions - CLEP Calculus
Card 0 of 10
On which of the following intervals is the function
continuous?
On which of the following intervals is the function
The function has a removable discontinuity at 
.
Since this function is undefined at 
is it not continuous across any interval containing 
.
Notice that the correct answer is an open interval that goes up to, but does not include 
.
The function has a removable discontinuity at
Since this function is undefined at
Notice that the correct answer is an open interval that goes up to, but does not include
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Describe the function
on the interval 
.
Describe the function
on the interval
This function (shown below) is defined for every value along the interval with the given conditions (in fact, it is defined for all real numbers), and is therefore continuous. However, there is a cusp point at (0, 0), and the function is therefore non-differentiable at that point.
This function (shown below) is defined for every value along the interval with the given conditions (in fact, it is defined for all real numbers), and is therefore continuous. However, there is a cusp point at (0, 0), and the function is therefore non-differentiable at that point.
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On what interval is the derivative of the function:
continuous?
On what interval is the derivative of the function:
continuous?
The derivative of the funtion using the power rule
is 
,
so it is not continuous when 
or is negative.
This occurs when 
, so the interval of continuity will be when 
, which is the interval 
.
The derivative of the funtion using the power rule
so it is not continuous when
This occurs when
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For what value(s) of 
?
For what value(s) of
For rational expressions, points of discontinuity often occur when the denomenator equals zero (triggering a division by zero).
Here we note that the denomenator will equal zero at positive and negative 4; however, we first must factor ensure that neither of these are factorable.
When factoring the rational expression the following occurs:
We now see that the point x=4 was a removable discontinuity and in fact the only time this function is discontinous is at x=-4.
Recall: For a function to be continous at a point, the point must exist, the limit must exist, and the limit must equal the point.
For rational expressions, points of discontinuity often occur when the denomenator equals zero (triggering a division by zero).
Here we note that the denomenator will equal zero at positive and negative 4; however, we first must factor ensure that neither of these are factorable.
When factoring the rational expression the following occurs:
We now see that the point x=4 was a removable discontinuity and in fact the only time this function is discontinous is at x=-4.
Recall: For a function to be continous at a point, the point must exist, the limit must exist, and the limit must equal the point.
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Evaluate the following limit:
Evaluate the following limit:
The limit as any asymptotic function (a function with an asymptote) approaches infinity is always its horizontal asymptote.
To find the horizontal asymptote of a rational expression one must first look for the following conditions:
- If the highest power of x is in the numerator then the horizontal asymptote does not exist.
- If the highest power of x is in the denomenator then the horizontal asymptote is at 0.
- If the highest power is in both, then divide the leading coefficents.
In the case of this problem the highest power is in both so we divide the leading coefficents
Therefore, our limit is 2.
The limit as any asymptotic function (a function with an asymptote) approaches infinity is always its horizontal asymptote.
To find the horizontal asymptote of a rational expression one must first look for the following conditions:
- If the highest power of x is in the numerator then the horizontal asymptote does not exist.
- If the highest power of x is in the denomenator then the horizontal asymptote is at 0.
- If the highest power is in both, then divide the leading coefficents.
In the case of this problem the highest power is in both so we divide the leading coefficents
Therefore, our limit is 2.
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On which of the following intervals is the function continuous?
On which of the following intervals is the function
The function has a removable discontinuity at .
Since this function is undefined at is it not continuous across any interval containing .
Notice that the correct answer is an open interval that goes up to, but does not include .
The function has a removable discontinuity at
Since this function is undefined at
Notice that the correct answer is an open interval that goes up to, but does not include
Compare your answer with the correct one above
Describe the function
on the interval .
Describe the function
on the interval
This function (shown below) is defined for every value along the interval with the given conditions (in fact, it is defined for all real numbers), and is therefore continuous. However, there is a cusp point at (0, 0), and the function is therefore non-differentiable at that point.
This function (shown below) is defined for every value along the interval with the given conditions (in fact, it is defined for all real numbers), and is therefore continuous. However, there is a cusp point at (0, 0), and the function is therefore non-differentiable at that point.
Compare your answer with the correct one above
On what interval is the derivative of the function:
continuous?
On what interval is the derivative of the function:
continuous?
The derivative of the funtion using the power rule
is ,
so it is not continuous when or is negative.
This occurs when , so the interval of continuity will be when , which is the interval .
The derivative of the funtion using the power rule
so it is not continuous when
This occurs when
Compare your answer with the correct one above
For what value(s) of
?
For what value(s) of
For rational expressions, points of discontinuity often occur when the denomenator equals zero (triggering a division by zero).
Here we note that the denomenator will equal zero at positive and negative 4; however, we first must factor ensure that neither of these are factorable.
When factoring the rational expression the following occurs:
We now see that the point x=4 was a removable discontinuity and in fact the only time this function is discontinous is at x=-4.
Recall: For a function to be continous at a point, the point must exist, the limit must exist, and the limit must equal the point.
For rational expressions, points of discontinuity often occur when the denomenator equals zero (triggering a division by zero).
Here we note that the denomenator will equal zero at positive and negative 4; however, we first must factor ensure that neither of these are factorable.
When factoring the rational expression the following occurs:
We now see that the point x=4 was a removable discontinuity and in fact the only time this function is discontinous is at x=-4.
Recall: For a function to be continous at a point, the point must exist, the limit must exist, and the limit must equal the point.
Compare your answer with the correct one above
Evaluate the following limit:
Evaluate the following limit:
The limit as any asymptotic function (a function with an asymptote) approaches infinity is always its horizontal asymptote.
To find the horizontal asymptote of a rational expression one must first look for the following conditions:
- If the highest power of x is in the numerator then the horizontal asymptote does not exist.
- If the highest power of x is in the denomenator then the horizontal asymptote is at 0.
- If the highest power is in both, then divide the leading coefficents.
In the case of this problem the highest power is in both so we divide the leading coefficents
Therefore, our limit is 2.
The limit as any asymptotic function (a function with an asymptote) approaches infinity is always its horizontal asymptote.
To find the horizontal asymptote of a rational expression one must first look for the following conditions:
- If the highest power of x is in the numerator then the horizontal asymptote does not exist.
- If the highest power of x is in the denomenator then the horizontal asymptote is at 0.
- If the highest power is in both, then divide the leading coefficents.
In the case of this problem the highest power is in both so we divide the leading coefficents
Therefore, our limit is 2.
Compare your answer with the correct one above