Sketch the graph of a function with the given properties. You ...

Question

Sketch the graph of a function with the given properties. You do not need to find a formula for the function.

p(0) = 2,lim x→0 p(x) = 0,lim x→2 p(x) does not exist, p(2)=lim x→2^+ p(x)=1

Accepted Answer

Welcome back, everyone. We want to identify the graph of the function with the given properties that F of 0 equals 0. The limit of F of X as X approaches 0 equals -5. The limit of F of X as X approaches 3 does not exist, and that F of 3 equals the limit of F of X as X approaches 3 from the right, which is 2. For our answer choices, we have the graphs in which we're going to look at each of our functions and then see which one of these match those properties. So let's go back to answer choice A, OK? And I think it would probably be helpful for me to just keep a copy of these functions as I go to each graph. Now, starting with A, is F of 0 equal to 0? We'll notice here that F of 0 instead. Is equal to 5. In other words, when X equals 0 up on our on our graph, F of X or Y is equal to, sorry, not 5, -5. So since F of 0 is -5 and not 0, then for the rest of properties, or rather we know that the answer choice cannot be a because since it doesn't match one of the properties, then it doesn't matter if it matches the rest. We're looking for the graph that matches all of the properties. Let's go on to answer choice be. Now, again, let's put our statement here, OK? Our properties, and let's start with F of 0. Is F of 0 equal to 0? Well, again, no. Notice that when X equals 0, F of X equals -2. So since F of 0 is -2 and not 0, B cannot be the correct answer. Let's move on to answer choice C, OK. Now for our answer try see and let me put the properties in the corner here. Let's start with F of 0. Is F of 0 equal to 0? Well, notice that when X equals 0, OK, then F of X is also equal to 0 because here is where our shaded or filled point is. Therefore, F of 0 equals 0. It matches that property. Next, let's check if the limit of F of X as X approaches 0 equals -5. To find that limit, we can look at the behavior of the function as X approaches 0 from the left and right. Notice that as X approaches 0 from the left, FF X approaches 55, sorry, and as it approaches 0 from the right, FFX, or sorry, FFX also approaches -5, OK. So as X approaches 0 from the left. F of X approaches -5. From the right, sorry, and as x approaches 0 from the left, F of X approaches -5. Thus this tells us that the limit as X approaches 0 of FX, since the left hand limit and the right hand limit are the same, it exists and is equal to -5. Thus it satisfies the second criteria. How about the third? Does the limit as X approaches 3 of FX exist? Well, when X equals 3, notice that from the right, OK, or sorry, from the left, I really need to get my directions properly sorted out. But notice that the limit as X approaches 3 from the left is equal to 4. And as X approaches through from the right, OK. The limit as X approaches 3 from the right of F of X is equal to 2. So no, because the left hand limit is not equal to the left to the right hand limit. Sorry. That means the limit of FF X as X approaches 3 does not exist. It satisfies that criteria. Let's go on to the final one. Is F of 3 and the limit of FFX as X approaches 3 from the right equal to 2? Well, we know that the limit as X approaches 3 from the right of FF X is equal to 2, but how about F of 3? Well, when our input is free, if we go up to our output, OK. And we come across here, we can tell that Fox 3. OK, F of 3 is also equal to 2. So both the limit of FX as X approaches 3 from the right and F of 3 are equal to 2. Therefore, since it satisfies all the criteria, C is the correct answer. Thanks for watching everyone. I hope this video helped.

Sketch the graph of a function with the given properties. You do not need to find a formula for the function.

p(0) = 2,lim x→0 p(x) = 0,lim x→2 p(x) does not exist, p(2)=lim x→2^+ p(x)=1

Key Concepts

Limits

Continuity

Piecewise Functions

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