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Numerade Educator

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Problem 51 Easy Difficulty

Let
$ B(t) = \left\{
\begin{array}{ll}
4 - \frac{1}{2}t & \mbox{if $ t < 2 $}\\
\sqrt{t + c} & \mbox{if $ t \ge 2 $}
\end{array} \right.$
Find the value of $ c $ so that $ \displaystyle \lim_{t \to 2}B(t) $ exists.

Answer

$c=7$

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Video Transcript

This is a problem. Number fifty one of the sewer calculus eighth edition section two point three Let the function B of thine equal. This piece was function where from For the values tea is listen to, the function is four minus one half of T and for the range or the domain of tea is greater than equal to to the function is the square root of the quantity T plus C Find the value of C so that the limit as t purchased two of the function be exists for this problem. We need to understand that this limit exists on ly if and only if the limit ist he approaches to from the left of B equals the limit is t approaches to from the right for being to happens, t equals to happens to be the splitting point between this function and dysfunction in the piece wise ah, function. So what we're gonna do is we're going to take the LTD's he approaches to from the left. How's it the function of B to the left of two, which is the first function for minus one, have team. And for the second limit as tempers tooth in the right. We are in this domain, which means that the function B is the second function the square root of the quantity t plus C. Now, in order for this limit to be true, these two have to be equal. And we are working our way towards figuring out what see value makes that true in this next deputies direct substitution too, since we have no domain restrictions And through that we get four minus one half times two, which is what he is a protein A is equal to for the second are for the right hand side. We have the square root of T equals two plus scene. Here we have a four minus one of terms to one of them to his one four minus one is three equals the square root of tipple scene In order to solve for C, we need to square both sides. So we do that you get nine on the left hand side equals to proceed and we see that see must be equal to seven. In order for the limits to exist, the limit is to purchase two of the function being Oh, and so we have solved for the value of see that it makes this true