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Find the values of π which make the function π continuous at π₯ equals π, if: π of π₯ is equal to two plus π₯ squared, if π₯ is less than or equal to π; and negative three π₯, if π₯ is greater than π.
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For a function π of π₯ to be continuous at π₯ equals π, three things must be true.
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Firstly, π of π must exist, that is the function π of π₯ is defined at π₯ equals π.
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Secondly, the limit as π₯ tends to π of π of π₯ must exist, so both the left-hand and right-hand limits exist and are equal.
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And thirdly, the limit as π₯ tends to π of π of π₯ must equal π of π.
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Letβs go through these one by one.
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First of all, we need π of π to exist.
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Well, when π₯ is less than or equal to π, which includes when π₯ is equal to π, π of π₯ is equal to two plus π₯ squared.
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So π of π is equal to two plus π squared.
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That is defined for any real value of π, so our first criterion is satisfied.
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Now we move onto making sure that the limit as π₯ tends to π of π of π₯ exists.
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First, we evaluate the left-hand limit, the limit as π₯ tends to π from below of π of π₯.
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For this one-sided limit, weβre only considering values of π₯ for which π₯ is less than π, and so π of π₯ is equal to two plus π₯ squared.
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And we evaluate this limit by direct substitution, replacing π₯ by π to get two plus π squared.
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Now we have to find the value of the other one-sided limit, the limit as π₯ tends to π from above of π of π₯.
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When π₯ is greater than π, π of π₯ is equal to negative three π₯.
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So this is the limit as π₯ tends to π of negative three π₯, which is negative three π.
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We have shown that both one-sided limits exist, but for the limit as π₯ tends to π period of π of π₯ to exist, the values of these two one-sided limits have to be equal.
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Two plus π squared has to be equal to negative three π.
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Letβs clear some room and solve this equation.
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First, we add three π to both sides and rearrange the terms.
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Now we have a quadratic in π in the form that weβre used to solving.
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We can factor this quadratic by inspection, and hence we see that these solutions are π equals negative one and π equals negative two.
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Of course, we couldβve used a different method, maybe completing the square or applying the quadratic formula.
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We wouldβve got the same solutions.
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These are the two values of π, for which the limit as π₯ tends to π of π of π₯ will exist.
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The only thing left to check is that the limit as π₯ tends to π of π of π₯ is equal to π of π.
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Recall that two plus c squared was the limit as π₯ tends to π from below of π of π₯ and negative three π was the limit as π₯ tends to π from above of π of π₯.
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We showed these two were equal when the π is equal to negative one or negative two.
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And when they are equal, it makes sense to talk about the limit as π₯ tends to π of π of π₯.
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Letβs first try π equals negative one.
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In this case, the limit as π₯ tends to π from below of π of π₯ is two plus negative one squared by direct substitution, which is three.
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We also do direct substitution to find the limit as π₯ tends to π from above, getting a negative three times negative one, which again is three.
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And as these two one-sided limits agree, we can say that the limit as π₯ tends to π of π of π₯ is three.
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To satisfy the third criterion, this limit must be equal to the value of the function at π.
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π of π is two plus π squared, so when π is negative one, π of π is two plus negative one squared, which is three.
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And we can see that the third criterion is satisfied, the limit as π₯ tends to π of π of π is equal to π of π when π is negative one.
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Now we just have to check π equals negative two.
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When π is equal to negative two, as we know that the values of the two one-sided limits are equal, the limit as π₯ tends to π of π of π₯ is just the value of one of the one-sided limits.
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Weβll choose the limit as π₯ tends to π from below of π of π₯ which is two plus π squared.
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Substituting the value of π, negative two in, we get two plus negative two squared, which is six.
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And of course, you can check this is exactly what we would get by direct substitution into the other one-sided limit.
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Whatβs the value of π of π in this case?
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Well, π of π is also equal to two plus π squared.
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And substituting negative two, we get a value of six.
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And again, the third criterion is satisfied.
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The limit as π₯ tends to π of π of π₯ is equal to π of π; theyβre both equal to six when π is equal to negative two.
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And so in this case, the limit as π₯ tends to π of π of π₯ is equal to π of π whenever the limit as π₯ tends to π of π of π₯ exists, i.e., when π is equal to negative one or π is equal to negative two.
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So there are two values of π which make the function π continuous at π₯ equals π, namely π equals negative one and π equals negative two.