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Numerade Educator



Problem 7 Easy Difficulty

Sketch the graph of a function $ f $ that is continuous except for the stated discontinuity.

Removable discontinuity at 3, jump discontinuity at 5.


The graph of $y=f(x)$ must have a removable
discontinuity (a hole) at $x=3$ and a jump discontinuity
at $x=5$.


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Video Transcript

Okay, we want to sketch the graph of a function F of X. We want a removable dis continuity at three. And the jump just continuity at five. A removable dis continuity when x is straight. So let's have an open circle right here, at X equals tree. And here is the graph of our function As X is approaching three from the left side here is the graph of the function um as X is approaching three from the positive side. Now, let's define a function when X is three to be this point right here. All right. So this is what we would call a removable dis continuity at three. You can see that this function is discontinuous at three because as the function approach, as X approaches three, okay, to function looks like a once approach approach this point. This value here instead, the function is defined to be this value up here. Uh This is a removable dis continuity. If we redefined ffx when X equals three to be this point. If we move this point of the function and put it here, uh then we would have a continuous function. So that's why this is called a removable dis continuity. A jump. This continuity at five is a little more serious. Let's suppose. Um as x approaches five from the left, uh the function is approaching uh this point right here. So this will be the value of the functions when X equals five, but we're going to continue our function um up here with an open circle because if uh F five is this point here, then we can't have another point of the function for the same X value five. So, open circle meaning this is not really on the graph, but then from here on for X is greater than five. We're continuing to graph uh over there. So now that we have an idea what the graph looks like, let's look at it. We said that at X equals three. We have removable dis continuity. The function is discontinuous. You have a low breaking a function but this discount this this this continuity is removable because we can redefine f of X when X equals three. Instead of being this point, put it down here and it will be continuous here. We have what's called a jump dis continuity. There's no way to remedy this. Um here, as X approaches five, the function approaches this point here, but then as we continue on past five, uh function continues on on this portion of the graph up here. Clearly, the function is discontinued. Clearly, the function is discontinuous at X equals five and this is called a jump discontinuity because the graph of F. Quebec's jumped from here and then up to here to continue