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

- [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.