Interpreting behavior of _ from graph of _'=Ä | AP Calculus AB | Khan Academy

Let g of x be equal to the definite integral from 0 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 5 to 10?

So, concave up! 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 introduce this variable t here is that this thing right over here is actually a function of x. Because x is this upper bound, it would be weird if we had x as an upper bound and we also, or at least confusing, and we're also integrating with respect to x. So, we just have to pick kind of another placeholder variable.

It 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 and this is y 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 open interval from 5 to 10. We have g's derivative graphed here, and we want to 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 the tangent line, or tangent slope of tangent, is increasing. Or another way of thinking about it, your derivative derivative is increasing. Another way to think about it—if your derivative is increasing over an interval, then you're concave up on that interval.

Here, we have a graph of the derivative, and it is indeed increasing over that interval. So, our calculus-based justification that we'd want to 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 sufficient. 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 are 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. 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 8?

So once again they've graphed f here, which is the same thing as the derivative of g. If we have the graph of the derivative, how can we say—how do we know that we have a relative minimum at x equals 8? Well, the fact that we cross, that we’re at the x-axis, that y is equal to zero, that the function, that the derivative is equal to zero at x equals 8, 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, goes from decreasing to increasing, and so you would have a relative minimum point.

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 8 and positive after x equals 8. That’s exactly what we just described.

Let's see about these. F is concave up on the interval around x equals 6—well, x equals 6 is a little unrelated to that. There’s an interval in the graph of g around x equals 8 where g of 8 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.

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 an appropriate calculus-based justification for the fact that g is positive on the closed interval from 7 to 12?

So the positive on the closed interval from 7 to 12—this is interesting! Let's just remind ourselves here; we're going to think a little bit deeper about what it means to be this definite integral from 0 to x. So as we add—if we think about what happens when x is equal to 7—when x is equal to 7, or another way of thinking about it, g of 7 is going to be the integral from zero to seven of f of t dt.

And so the integral from 0 to 7—if this was the 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 0 to 7, 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 7 to 12, we're not adding any more area, but we’re also not taking any away.

So actually, g of 7 all the way to g of 12 is going to be the same positive value because we're not adding any more value. When I say g of 12, g of 12 is going to be actually equal to g of 7 because once again, no added area right here, positive or negative.

So let's see which of these choices match: for any x value in the interval from 7 to 12, the value of f of x is 0. 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 have 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 7 to 12, the closed interval, the value of g of x is positive. For any x value in the interval from 7 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 0 to 7, and it is non-negative over 7 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 0 to 7, so it develops all this positive area, and it's non-negative over the interval. So, we're going to stay positive this entire time for g, which is the area above under f and above the x-axis from 0 to our whatever x we want to pick.

So I like this choice here: f is neither concave up nor concave down over the interval from the closed interval from 7 to 12. No, that doesn't really help us in saying that g is positive over that interval, so there you go, choice c.