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AP Calc: FUN‑5 (EU), FUN‑5.A (LO), FUN‑5.A.3 (EK)

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- [Instructor] Let g of x be equal to the definite integral from zero to x of f of t dt. What is an appropriate calculus-based justification for the fact that g is concave up on the open interval from five to 10? So concave up. So before I even think about what it means to be concave up, let's just make sure we understand this relationship between g and f. One way to understand it is if we took the derivative of both sides of this equation, we would get that g prime of x is equal to f of x. The derivative of this with respect to x would just be f of x. In fact, the whole reason why we introduced this variable t here is this thing right over here is actually a function of x 'cause x is this upper bound. And it would've been weird if we had x as an upper bound, or at least confusing, and we were also integrating with respect to x. So we just had to pick kind of another placeholder variable. Didn't have to be t. It could be alpha, it could be gamma, it could be a, b, or c, whatever we choose, but this is still, right over here, this is a function of x. But when you take the derivative of both sides, you realize that the function f, which is graphed here. And if this were the x axis, then this would be f of x. If this is the t axis, then this is y is equal to f of t. But generally this is the graph of our function f, which you could also view as the graph of g prime. If this is x, this would be g prime of x. And so we're thinking about the interval, the open interval from five to 10, and we have g's derivative graphed here. And we wanna know a calculus-based justification from this graph that lets us know that g is concave up. So what does it mean to be concave up? Well, that means that your slope of tangent line, of tangent, slope of tangent is increasing. Or another way of thinking about it, your derivative is increasing. Or another way to think about it, if your derivative is increasing over an interval, then you're concave up on that interval. And so here we have a graph of the derivative, and it is indeed increasing over that interval. So our calculus-based justification that we'd wanna use is that, look, f, which is g prime, is increasing on that interval. The derivative is increasing on that interval, which means that the original function is concave up. f is positive on that interval. That's not a sufficient calculus-based justification. Because if your derivative is positive, that just means your original function is increasing. It doesn't tell you that your original function is concave up. f is concave up on the interval. Well, just because your derivative is concave up doesn't mean that your original function is concave up. In fact, you could have a situation like this where you're concave up over that interval, but for much of that interval right over here, if this was our graph of f or g prime, we are decreasing. And if we're decreasing over much of that interval, then actually on this part our original function would be concave down. The graph of g has a cup U shape on the interval. Well, if we had the graph of g, this would be a justification, but it wouldn't be a calculus-based justification. Let's do more of these. So this next one says, so we have the exact same setup, which actually all of these examples will have. g of x is equal to this thing here. What is an appropriate calculus-based justification for the fact that g has a relative minimum at x equals eight? So once again, they've graphed f here, which is the same thing as the derivative of g. And so if we have the graph of the derivative, how do we know that we have a relative minimum at x equals eight? Well, the fact that we cross, that we're at the x-axis, that y is equal to zero, that the derivative is equal to zero at x equals eight, that tells us that the slope of the tangent line of g at that point is zero. But that alone does not tell us we have a relative minimum point. In order to have a relative minimum point, our derivative has to cross from being negative to positive. Why is that valuable? Because think about if your derivative goes from being negative to positive, that means your original function goes from decreasing to increasing. It goes from decreasing to increasing. And so you would have a relative minimum point. And the choice that describes that, this is starting to get there, but this alone isn't enough for a relative minimum point. f is negative before x equals eight and positive after x equals eight. That's exactly what we just described. Let's see about these. f is concave up on the interval around x equals six. Well, x equals six is a little bit unrelated to that. There's an interval in the graph of g around x equals eight where g of eight is the smallest value. Well, this would be a justification for a relative minimum, but it is not calculus-based. So once again I'll rule that one out as well. Let's do one more of these. So same setup, although we have a different f and g here, and we see it every time with the graph. What is a appropriate calculus-based justification for the fact that g is positive on the interval from the closed interval from seven to 12? So the positive on the closed interval from seven to 12. So this is interesting. Let's just remind ourselves. Here we're gonna think a little bit deeper about what it means to be this definite integral from zero to x. So, if we think about what happens when x is equal to seven. When x is equal to seven, or another way to think about it, g of seven is going to be the integral from zero to seven of f of t dt. And so the integral from zero to seven, if this was a t-axis, and, once again, t is just kind of a placeholder variable to help us keep this x up here. But we're really talking about this area right over here. And because from zero to seven this function is above the x-axis, this is going to be a positive area. This is a positive area. And as we go from seven to 12, we're not adding anymore area, but we're also not taking any away. So actually g of seven all the way to g of 12 is going to be the same positive value, 'cause we're not adding anymore value. And when I say g of 12, g of 12 is going to be actually equal to g of seven, because, once again, no added area right here, positive or negative. So let's see which of these choices match. For an x value in the interval from seven to 12, the value of f of x is zero. That is true, but that doesn't mean that we were positive. For example, before that interval if our function did something like this, then we would've had negative area up to that point, and so these would be negative values, so I would rule that out. For any x value in the interval from seven to 12, the closed interval, the value of g of x is positive. For any x value in the interval from seven to 12, the value of g of x is positive. That is true, so I like this one. Let me see these other ones. f is positive over the closed interval from zero to seven, and it is non-negative over seven to 12. I like this one as well. And actually the reason why I would rule out this first one, this first one has nothing to do with the derivative and so it's not a calculus-based justification, so I would rule that one out. This one is good. This is the exact rationale that I was talking about. f is positive from zero to seven, so it develops all this positive area, and it's non-negative over the interval. And so we are going to stay positive this entire time for g, which is the area under f and above the x-axis from zero to our whatever x we wanna pick. So I like this choice here. f is neither concave up nor concave down over the closed interval from seven to 12. No, that doesn't really help us in saying that g is positive over that interval. So there you go, choice C.
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