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Discuss the continuity of the function π, given that π of π₯ is equal to three plus the sin of π₯ if π₯ is greater than or equal to zero and π₯ is less than π by two and π of π₯ is equal to four plus π₯ minus π by two to the eighth power if π₯ is greater than or equal to π by two.
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The question wants us to discuss the continuity of a function π.
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And we can see that this function is defined piecewise.
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To discuss the continuity of our function π, letβs recall what it means for our function to be continuous at a point π₯ is equal to π.
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We say π is continuous at π₯ is equal to π if π evaluated at π is equal to the limit as π₯ approaches π of π of π₯.
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And, of course, both π evaluated at π and this limit must exist.
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Because weβre dealing with the continuity of a piecewise-defined function, we can check this in the following three steps.
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First, we want to find the domain of our function π of π₯, since a function can only be continuous on its domain.
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Second, we check the continuity on each interval of our piecewise-defined function.
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Finally, we just need to check that the endpoints of these intervals, our functions, match.
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Letβs start by checking the domain of our function π of π₯.
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To do this, we need to check the values of π₯ for which our piecewise-defined function is defined.
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We see when π₯ is greater than or equal to zero and π₯ is less than π by two, our function π of π₯ is equal to three plus the sin of π₯.
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And three plus the sin of π₯ has a domain of the entire set of real numbers.
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So in particular, this means itβs defined for all values of π₯ greater than or equal to zero and less than π by two.
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So our function π of π₯ is defined in this interval.
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Next, we can do the same when π₯ is greater than or equal to π by two.
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We get π of π₯ is equal to four plus π₯ minus π over two to the eighth power.
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And this is a polynomial.
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So its domain is, again, the entire set of real numbers.
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And, in particular, this means itβs defined for all values of π₯ greater than or equal to π by two.
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So weβve shown our function π of π₯ is defined on both of these intervals.
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So the domain of our function π of π₯ is both of these intervals combined.
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Thatβs π₯ greater than or equal to zero.
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We now want to check the continuity of our function π of π₯ on each of these intervals.
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Letβs start by checking the continuity for π₯ greater than or equal to zero and π₯ less than π by two.
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On this interval, our function π of π₯ is equal to three plus the sin of π₯.
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And we know the sin of π₯ is continuous on the set of real numbers because all trig functions are continuous on their domains and the sin of π₯ is defined for all real numbers.
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Next, constant functions are just polynomials.
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So theyβre continuous on the set of real numbers.
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Finally, this means three plus the sin of π₯ is the sum of two continuous functions, which means that itβs continuous.
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And this means our function π of π₯ is continuous over this interval.
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However, we must be careful to remove any endpoints where our function π of π₯ changes definition.
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We can now do the same for values of π₯ greater than or equal to π by two.
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We can see, for these values of π₯, π of π₯ is a polynomial.
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And, of course, we know all polynomials are continuous on the set of real numbers.
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And we must be careful at this point, since this only tells us our function π of π₯ is continuous when π₯ is strictly greater than π by two.
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Since our function π of π₯ changes definition when π₯ is less than π by two and when π₯ is greater than π by two.
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So we now have our third and final step.
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We need to check that the endpoints of our function π of π₯ match.
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And we can see the only endpoint where our function changes definition is at π₯ is equal to π by two.
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Letβs start with the endpoint of three plus the sin of π₯ when π₯ is less than π by two.
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This is just a constant plus a trigonometric function.
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So we can do this by direct substitution.
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We get three plus the sin of π by two, which we can evaluate to just give us four.
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The endpoint of our second interval is even easier.
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Since π₯ must be greater than or equal to π by two, our endpoint would just be where π₯ is equal to π by two.
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So we just directly substitute π₯ is equal to π by two into four plus π₯ minus π by two to the eighth power.
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This gives us four plus π by two minus π by two to the eighth power, which we can evaluate to give us four.
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And we can see, in this case, that these endpoints match.
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Theyβre both equal to four.
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So this tells us that π is continuous at π₯ is equal to π by two.
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So when we check the entire domain of our function π of π₯, we actually show that π of π₯ was continuous on its entire domain.
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Therefore, weβve shown that the function π of π₯, which is equal to three plus the sin of π₯ if π₯ is greater than or equal to zero and π₯ is less than π by two.
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And π of π₯ is equal to four plus π₯ minus π by two to the eighth power if π₯ is greater than or equal to π by two.
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Is continuous for all values of π₯ greater than or equal to zero.