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FUN‑5 (EU)

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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 interval the open interval from 5 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 because X is this upper bound and it'll been too weird if we had X as an upper bound and we also are at least confusing and we're also integrating with respect to X so we just have 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 we need to 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 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 from the open interval from 5 to 10 and 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 it doesn't 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 derivative is increasing or another way to think about if your derivative is increasing over an interval then you're concave up on that interval and so we here we have a graph of the derivative and it is indeed increasing over that end 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 base 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 you 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 set up 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 and so 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 if the fact that we cross that we're at the x axis that x equals that y is equal to zero that the function 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 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 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 bit unrelated to that and there's an interval in the graph of G around x equals 8 where G of 8 is a smallest value well this would be a justification for relative minimum but it is not calculus based so what's gal I'll rule that one out as well let's do one more of these so same set up 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 interval from the closed interval from 7 to 12 so the positive on the closed interval from 7 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 0 to X so as we at if we take if we think about what happens when X is equal to 7 when X is equal to 7 or another way to think about it G of 7 is going to be the integral from 0 to 7 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 close interval the value of G of X is positive for any value 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 zero 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 and so we are going to stay positive this entire time for G which is the area above under f and a bluff above the x-axis from 0 to R 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