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Toeul LuIf possible, find the number k S0 that f iS continuous atevery point |1ox+ 2, X< - 3 f(x) = KX + 7 X2 -30A: k=0 B, K=CC. K=D D. Tnere IS nc such value 0&...

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Toeul LuIf possible, find the number k S0 that f iS continuous atevery point |1ox+ 2, X< - 3 f(x) = KX + 7 X2 -30A: k=0 B, K=CC. K=D D. Tnere IS nc such value 0" kselectvour ans €3eC '

Toeul Lu If possible, find the number k S0 that f iS continuous atevery point |1ox+ 2, X< - 3 f(x) = KX + 7 X2 -3 0A: k= 0 B, K= CC. K= D D. Tnere IS nc such value 0" k selectvour ans € 3e C '



Answers

Find a value for the constant $k$, if possible. that will make the function continuous. (a) $f(x)=\left\{\begin{array}{ll}7 x-2, & x \leq 1 \\ k x^{2}, & x>1\end{array}\right.$ (b) $f(x)=\left\{\begin{array}{ll}k x^{2}, & x \leq 2 \\ 2 x+k, & x>2\end{array}\right.$

We want another values of K that will make these functions continuous for the first one. The only possible point where there could be discontinuity Is that the X value of one, since that's where changes from one function to the next. In order for it to be continuous, we have to have the limit as X approaches one from the negative side and his ex approaches one from the positive side. Be equal to each other if it's one from the negative side, we're looking at the top definition of seven X minus two one. From the positive side, it's the bottom definition K X square. If I fill in one for X, the first one is five. The second one would be K Times one or K. The two values must equal to make this continuous. If it's OK, it's five for our second one. The only possible break is at two. Let's first look at the limit as X approaches two from the negative side where we would use the top definition. And then let's look at the LTD's X approaches to from the positive side, which would be the bottom definition on the top one. If you fell to infer acts, you will get four. Okay, The bottom one. You will get four plus k. Now we need those two to be equal to each other to make this continuous. So let's set them equal. Subtract K. That would give us three k equals four. And if we divide by three, que is four third's.

Buddy Ricky and today, working on from 37 from the diagnosis, were given a function, and we need to determine for what value of K is continuous. So let's start with our three X squared minus three. Thanks for this one. And that's factor the top city. Probable. Three. Get next. Nice warm, that's all. Oh, wait, can Simply falling left? Three. Now we want to take the limit as ex churches, one in salt. And so what we get is that as X approaches one, we started six and you could determine this by looking and values that are super close to 1.9 council later come out to an answer that's possibly six, so that's video is helpful.

We're trying to find the K values that make this function continuous. For the first one, the bottom function would be discontinuous effect zero. However, we put a limitation that X is less than negative. Three. That means the only possible point where this could be discontinuous, as at the point where it's changing its definition because of it being a piece wise function. So let's examine the limit Is X approaches negative three from both the positive and negative side is X approaches negative. Three. From the negative side, that's values a little bit less than negative. Three. So let's fill in K over X Square. Positive side would be for X values a little bit greater the negative three. So that would be nine minus x squared in the first one. If I fill in a negative three, I would have K overnight the bottom one, I would have nine minus nine or zero. If those two values are the same, then I have a continuous function. Multiply by nine to make this continuous at all points. Okay, Must be zero. Be a similar, uh, this time again, it would be discontinuous effect zero for or the bottom function, but we have excesses strictly less than zero. So what's examine this As X approaches zero from the negative side and his ex approaches zero from the positive side. From the negative side, I'm going to use the bottom definition K over X square. And from the positive side is the top definition nine minus X. Where to limits must be equal If this is a continuous function on the bottom one. If I fill in zero for X, I get nine top when I get k over zero. Well, k over zero is not a real answer. So it doesn't matter what K over zero is equal to. There is no que value that will make this continuous.


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