The Numerator 12 X is Continuous on the Interval

�� CONTINUITY

The function, " �", is continuous at "a" if the limit at " �" exists AND equals the value of the function evaluated at " �".��

Recall from the previous week's lecture that for the limit to exist at " �", both the limit from the left and from the right of "a" must BOTH exist AND must be equal to each other. So, continuous at " �", implies all of this:��

We highlighted the basic continuity issues in the previous lecture. Now we'll explore some applications.

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Theorem : If " �" is continuous at "c" and

" �" is continuous at " �" , then the composite function composite function:

" �" is continuous at "c".

Example # 1 : Construct the composite function:

" �"; show that it is continuous at " �"; and graph all three functions.

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First show that " �" is continuous at " �" and that " �" is continuous at " �".

" �" is a polynomial and thus continuous on the interval: , which of course includes

" �".

" �" is a rational function and thus continuous everywhere on the interval: , except at

" �", where " �" has a vertical asymptote. So, " �" is continuous at " �".

Then by the previous theorem, the composite function: " �" is continuous at " �".

Here is the composite function.

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On the graph below, find " �" and then " �" and confirm that:

" �".

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Theorem : If " �" is continuous on a closed interval: [ a, b ] and "k" is any number between

" �" and " �" inclusive, then there is at least one number "c" in the interval: [ a, b ] such that

" �".�

This is the Intermediate Value Theorem (IVT ). Essentially, it guarantees us that every number between and including the "y" values at the endpoints of the closed interval will occur at least once on that closed interval provided that

" �" is continuous everywhere on the interval including at the interval's end values: "a" and "b".

The left end value of any closed interval cannot be approached from the left and the right end value can not be approached from the right. Therefore, when we say that " �" is continuous at the end values of the closed interval what that implies is this.��

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Example # 2 : Use the IVT to show that there is at least one root of the given polynomial on the specified interval and illustrate the results graphically.�

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Since polynomials are continuous on the open interval: �, then they must be continuous on every closed interval. The given interval is closed so the IVT applies.

Evaluate " �" at the end values of the interval: " " and " ".

A "root" of the function is a value of "x" where " �" is "0" and "0" lies between " �" and " �".

Therefore the IVT guarantees that there must at

least one value of "x" for which " �" is zero on the given interval.

Here is the graph of " �" over the given interval.

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Judging from the graph above, it appears as though there are two roots on the interval.

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Theorem : All of the Basic Trigonometric Functions are everywhere continuous in their respective domains.

Example # 3 : Find the discontinuities.�

This is one of the six basic Trigonometric Functions; therefore, it is continuous everywhere in its natural domain, which excludes: " �" for all integers "m" and zero.

Example # 4 : Find the discontinuities.�

This is one of the six basic Trigonometric Functions; therefore, it is continuous everywhere in its natural domain, which includes all real numbers. This sine function contains a polynomial in its argument. Since the range of every polynomial is a subset of the set of all real numbers, " �", is everywhere continuous.

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Definition : A function is continuous on a closed interval: [a,b] if it is continuous on the open interval: (a,b) and it is continuous from the right at "a" and it is continuous from the left at "b".

Example # 5 : Determine where "" is continuous.�

The polynomial under the radical must be non-negative.

Thus, the natural domain of this function is the closed interval: [,].

On the open interval: , we obtain this result

for any "c" on .

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Therefore, " �" is continuous on the open interval.

For the interval end values we obtain these results.

Therefore, " �" is continuous at interval's end values.

Thus, " �" is continuous on the closed

interval: [,].�

Theorem : If " �" is a one-to-one function that is continuous at each point of its domain, then

" �" is continuous at each point of its domain, that is " �" is continuous at each point in the range of " �".

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Example # 6 : Determine where "" is continuous.�

The fraction will be continuous at all points where the numerator and denominator are both continuous and the denominator is non-zero.

The domain of the function: " �" is . The range of the function:" �" is the domain of its inverse: " �", which is . We must therefore restrict "x" to the interval: �so that the range of " �" is .� Thus, the numerator is continuous on the open

interval: .

The denominator is a polynomial so it is continuous everywhere; but the ratio is undefined at .

Thus " �" is continuous on the interval:.

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Theorem : If �for all "x" in some open interval containing "c" and , then .�

Example # 7 : Prove that:.�

Examine the following Figure.

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There are three areas of interest here. The first is the area, " �", of the sector with the central angle, " x "; the second is the area, " �", of the triangle with these 3 vertices:, , ; and the third is the area, " �", of the triangle with these 3 vertices:, , .