2026 AP Calculus AB FRQ #4 Answers

Added: 2026-05-14

[00:00:00]All right. In the last couple videos, I explained FRQs 1, two, and three on the AP Calc exam from 2026. So, in this video, I'll explain number four from the calc AB test. We're told, let f be a twice differentiable function on the closed interval with f of 2 is 3. And we're told that this is the graph of f prime. And then we want to find for part a that the function g ofx is f ofx - ln of x. And we need to find g prime of 2.

[00:00:25]First of all, g prime of 2 will equal well actually let's just say g prime of x. So g prime is going to be equal to frime of x minus what is the derivative of ln of x? It's just 1 /x. And so g prime of 2 will equal f prime of 2 - 1 /2. And good thing we have frime of 2.

[00:00:47]It's 1.5, right? So it'll just be 1.5 minus one. It's just 1.5 minus 0.5 which is just one. Right? So the answer to that one is just one. Right? Okay. Let's go to the next one. We have find all values of x on the open interval at which the graph of f of just f has a point of inflection. When does f have a point of inflection? The point of inflection occurs when fp prime equals zero. Now, that also occurs when fp prime has a relative max or min.

[00:01:24]Relative max or min. And so here, what's going to happen is that I'm just looking for the the relative max or mins of f, right? And so it looks like we're going to have our point of inflection at -3 comma1, right? -3 comma 1 because it looks like well fprime fprime or sorry frime went from decreasing to increasing right and then we have uh here increasing to decreasing. So that's another point of inflection. And then we have frime increasing to decreasing as well, right? And so here we're going to have that it is at um -3 negative or sorry3 positive 1 and uh what three though because and then we could write out so like x equals this and then I could write out that uh does it say give a reason uh you could say these are the values where fprime has a relative extremma.

[00:02:44]which means that FP prime changes signs or something like that, right? And that means that we have a point of inflection. We could also say that frime changes from increasing to decreasing or decreasing to increasing, but um it doesn't really make a difference, right?

[00:03:04]Okay, continuing on to part C. On what open intervals, if any, if the is the graph of f both increasing and concave down? Well, when is the graph increasing? That's going to be when f is positive, right? So, it's going to be increasing. So, f is increasing on the interval of -2 2 3 u 3 2 4. Right? That is again when f prime is positive. And then when is it concave down? Well, concave down that is when my fp prime is going to be decreasing, right? Again, because that means that the derivative of that fp prime is less than zero, right? And so we're going to have so f concave down. That's going to be on the interval of -4 tog -3 and 1 2 3. Right?

[00:04:00]And so when is this satisfied? is going to be on just that interval of 1 2 3 right because that's where both of these uh situations are satisfied right I don't think that any other interval satisfies that and we could say does it say give a reason for your answer it does we could say something along the lines of negative 1 to three because this is where f prime is greater than zero uh meaning that f is increasing and also where fprime is decreasing meaning that fp prime is less than zero and therefore f is concave down right okay then the last part part d is asking us to find the absolute minimum and absolute maximum uh yeah absolute minimum and absolute maximum and so what we would do here is we have to use the candidates test. So we're going to use all of the end points. So x, by the way, how could we find f ofx? Well, f ofx, we're going to use the fundamental theorem of calculus.

[00:05:15]It's going to equal f of 2 plus the integral from 2 to we could say a of f of or frime of t dt. Right? If you search up the fundamental theorem of calculus, it's f of b= f of a plus the integral from a to b of frime of tdt.

[00:05:35]Right? Basically just plugging it into that. This is something that you should know uh before you take your exam, right? And so x and then we could just say f ofx, right? And so we're going to have here we actually might have to extend this page. So we're going to have that our end points here we have neg4 and then okay well when will f have an absolute minimum or maximum it's going to be at all these values where frime is equal to zero. So we'll try at -2 and we'll try at three and we'll try again at four. Um and then that should be good right and so then all we have to do is figure out then we could do the integral. So we could do like f of two. Did we find any of these values in previous answers? We had uh what? G prime. No, that doesn't help.

[00:06:25]Okay, so we have uh so f of -4 that's going to equal the uh that's going to equal three plus this area, right?

[00:06:39]We can't uh really tell this exact um value, right? However, what we do know is that from two to if we take the integral from 2 to -4, we're going to have this big. First of all, the integral is going to end up being negative, right? Because if I have negative So, if I have the integral from two to four of f of t, we have to flip it. It's the negative integral from4 to two of f of t. And originally like this integral should be something very like positive because this is a huge area on top here but it ends up being negative.

[00:07:17]So it'll just be like minus we could just say big amount right let's just leave it like that. Okay, I guess we shouldn't really say it like that, but then we have from -2 to -2, this value is going to be minus a small. So this will be like three minus it'll actually be a bigger amount. Okay, and the reason being that now we don't have this extra like value on bottom that usually it ends up being a negative integral, but we have to flip it and make it positive because we're going backwards. So currently this is going to be our max right if we're going from then two to three. So now from two to three this is going to be like three plus you know some small amounts right small amount and then from 2 to 4 that would be three plus we'd have this area plus this whole area. So big amount right? So it ends up being that we would just have um the absolute minimum at x= -2, right? And then for the absolute maximum, that would just end up being at x= 4, right? And I would explain it a little bit better on an FRQ, but I would say that the area from like -2 tog4 or sorry from -2 to 2 is bigger than the area from 4 to two because it doesn't include this negative part and so it ends up being a bigger change. And then I would say the same thing from like for three and four. The area from 2 to 3 is smaller than the area from two to four.

[00:08:58]So the four is is a bigger value. So x= -2 is the absolute minimum. X= 4 is the maximum.