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Problem 16 Easy Difficulty

Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval.

$ g(x) = \frac{x - 1}{3x + 6}, \hspace{5mm} (-\infty, -2) $

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Okay, we have a function G of x equals x minus 1/3. X plus six on the interval negative infinity to negative two. This is an open interval, not a closed interval at negative two. So that means the number negative two is not included in this interval. And that's good because uh if X was allowed to equal negative to um then three times negative two would be negative six and then plus six would be zero and you can't divide by zero But -2 is not in this interval. Uh so we do not have to worry about the denominator being zero anywhere on this interval. Now we are going to show uh that this function G of X is continuous on this interval. And uh for a function to be continuous, a function is continuous. G will be continuous at a some number A Okay, in this interval uh will show that it works for every number A indiscernible, but G is continuous at A If the limit of G of X as X approaches A is equal to G obey. This is what it means to have continuity. The limit of the function as X approaches a number A equals the value of that function at a. Yeah, let's go ahead and over here to decide. Let's find out what govs G of A. For any number A in this interval, Member, the only number in this is the only number that would give us trouble is negative two, but negative two is not in this interval. So for any number a G of a, simply plugged uh a in for XG of a would be a -1 Over three times a plus six. So that's gff. Now, if we could show that the limit of G of X as X approaches a equals G of a, then we will have shown that she is continuous on the interval. So what we want to do now is find a limit of G fx, which is x minus 1/3 X plus six. Find a limit of this expression as X approaches A. Well, The limit of X -1 as X approaches a exists, it's simply just a -1. And the limit of uh three times X plus six as X approaches A exists and is three times eight plus six. And so the limit of x minus 1/3, X plus six as X approaches A is going to equal to limit Of the Function and the Numerator X -1. As X approaches a divided by the limit of the function, the denominator, which is three X plus six as X approaches A. So in other words, the limit of a closure is equal to a close version of the limits. As long as the limit of the function in the denominator okay exists and is not zero Now, remember, the only number that would make three X plus six equal to zero is if X was equal to negative two and our number A is not going to be negative too because A is any number in this interval but negative two is not in this interval, so they will not be -2. And so we are allowed to take instead of the limit of the close ship. We can take the closure. And the fraction of the limits as long as the limit of the function that was in the denominator exists, which it will and is not zero, which of course won't be zero. So, now, what are these limits? Well, the limit Of X -1 as X approaches a is simply just a -1. We can just substitute it directly in there. The limit of the function three times X plus six as X approaches A is going to be three times A plus six. So the limit of our function G f X as X approaches A equals a minus 1/3 8 plus six, but G of A also equals a minus 1/3, 8 plus six. So we just showed for any A. In this interval the limit of G of X as X approaches A is equal to gov. So the limit of G of X as X approaches A equals G of A. So G is continuous at A because the limit of G of X as X approaches A does equal Gov.