Sketch a graph of f and use it to make a conjecture about the ...

Question

Sketch a graph of f and use it to make a conjecture about the values of f(a), lim x→a^−f(x),lim x→a^+f(x), and lim x→a f(x) or state that they do not exist.

f(x) = {√x if x<4

3 if x=4; a=4

x+1 if x>4

Accepted Answer

Welcome back, everyone. In this problem, we want to analyze the P function HX given by H X equals the root of X minus 2 if X is less than 63 if X equals 6, and X minus 2 if X is greater than 6, where A equals 6. Skech a graph of H and determine the values of H of A, the limit of H of X as X approaches A from the left as X approaches A from the right, and as X approaches A or state that they do not exist. For our answer choices, it shows the possible values and the possible graph, and in A it says that H of A is 4. The limit of H of X as X approaches A from the left is 2, as it approaches A from the right is 3, and as it approaches A, it does not exist. B says that there are 234, and that it does not exist, respectively. Here is our graph. She says that it's 3 to 4 and does not exist respectively, and here's our graph. And the D says it's 32 4 and it does not exist, and here is our graph. Now I've attached a graph here as well to help us graph H of X. So let's rewrite our function, put it on the graph, and then determine the possible values of the function for a given value of A and the limits as outlined in the problem statement. So we're given the function H of X. As a piecewise function. Where it's the root of X minus 2, OK. If X is less than 6, it's 3, if X equals 6, and it's X minus 2, if X is greater than 6, OK, where A equals 6. No, that means they're going to be 3 parts to our function. Let's start with the part that's where X sorry, is less than 6. No, 4 X less than 6. We know that H of X is going to be equal to the square root of X minus 2. Now, essentially this is a graph of root X shifted 2 units to the right. OK. So that means we're going to draw root X and shift our reference point from 00, 2 units to the right to get the new reference point at 20. And now we know this goes up to where X is less than 6, OK. And when X is 6. Then H X equals the root of 6 minus 2. So let's make a note of that here. When X equals 6, H of X equals the root of 6 minus 2.6 minus 2 is 4, and the root of 4 equals 2, OK? So it's going to go up to the.62, but it doesn't include the 0.62, OK? So here we have 62, but we know at this point there's going to be a hole. Cause it doesn't include that point. And now, let me put it in red here. We can draw our graph of H X where X is less than 6. Now, when X equals 6, OK, then H of X equals 3. So thus the 0.63 is going to be on our graph, OK? So we're going to go to 63 and now we know that this point is included, so let's draw a circuit here and let's shade it because this is going to be the 0.63, OK? And we know we have the 0.62 down here. Now, finally, OK, I feel like I made this a bit big. Let me try to shrink it a little bit. Can I? Uh, yes, I can. OK, just to make it a little smaller. And now finally, we know that for X greater than 6, OK? Then H of X is going to be equal to x minus 2. So now we know there's going to be a hole at X equals 6. And for this function, when X equals 6, OK, when X is 6, then 6 minus 2 equals 4, so that means 64 is a hole. OK. So we could start there, OK. And if we choose randomly some other value of, of, of X, for example, for example, let's say we use uh 9, when X is 9, H of X is 9 minus 2, which is 7. So 97 is a point on, so let me fix that here. On the graph, OK. There's a point on the graph and X is greater than 6, so we can go to 97. Oops, let's break it down a bit. Let's take it to 86, because my graph doesn't go up that far. So 86 is a point on the graph, so now we have our line and now we can draw X sorry, 86, OK. So now we know that our graph is going to look something like this, OK? And know that we have our graph, then we can start to figure out the values. In other words, no. We can try to find the value of. H of A, we know A is 6, OK. So since A equals 6. Now we can try to find H of 6, OK? We can find the limit of H of X. As X approaches A or 6 from the left, the limit as X approaches 6 from the right. And then the limit of H of X as X approaches 6. Now, on our graph, OK. Notice that 4 H of 6, OK. That means when X is 6, what is H? Well, it would be 3. So H of 6 equals 3. Next, to find the limit of H of X as X approaches 6 from the left, we have to look at the graph's behavior to the left of 6. Now notice that as X approaches 6, H of X approaches 2 from the left. Therefore, the limit is going to be equal to 2. Now, when we look at H of X as X approaches 6 from the right, OK. Notice that as X approaches 6 from the right, H of X approaches 4. So what that means then is that the limit as it approaches 6 from the right equals 4. Now, what is going to be the limit of H of X as X approaches 6? Well, this limit exists if the left hand limit equals the right-hand limit. In this case, we found the left-hand limit as X approaches 6 and it's right hand limit and they are not equal. OK. So since the left-hand limit does not equal the righthand limit, then the limit of H of X as X approaches 6 does not exist, OK? It does not exist because the left hand limit. Is not equal to the right hand limit. Therefore, For our answer choices that means that H of 6 is 3. The limit of H of X as X approaches 6 from the left is 2, as it approaches 6 from the right is 4, and the limit as X approaches 6 does not exist. Now if we go back to our answer choices and look at our graphs, then we can tell that D is the correct answer. Thanks for watching everyone. I hope this video helped.

Sketch a graph of f and use it to make a conjecture about the values of f(a), lim x→a^−f(x),lim x→a^+f(x), and lim x→a f(x) or state that they do not exist.
f(x) = {√x if x<4
3 if x=4; a=4
x+1 if x>4

Key Concepts

Piecewise Functions

Limits

Continuity

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