@JoshuaJonme Then what if (a, b) was the interval where f’(x)<0? You have to find something that’s true for all a and b such that 0<a<b<5.
@Javacash01 I used this example earlier, so I’ll try again. Consider f(x) = x^3 on the interval [-2, 1]. For every a and b such that -2<a<b<1, f(a)<f(b) for this function. However, f’(x) = 0 at x = 0.
@Javacash01 f(a) and f(b) can get infinitely close to each other but can NEVER BE THE SAME. If If f’=0 at a single point then f(b) is still greater than f(a). Like if f’(3)=0 and a=2.999999999 and b=3.0000001 f(b) > f(a) if the function increases everywhere else.
@RHSclassof16 @dogeater @Javacash01 Okay lets not be uncivilized. Both answers make sense, the wording of the problem most likely decided which was correct. Move on to a new question.
@dogeater is correct.
@dogeater is right… the answer for that one was f’(x) > 0 … the graph of f(x) is always increasing on that interval because f(b) > f(a) and b > a therefore the slope will always be positive.
For the FRQ, part d for one of them seemed to be impossible… the one where it asks for y=mx + b as a solution for the differential equation.
Was one of the answers for an MC calculator question 7/9? it asked for the integral of (f(x))^2 and said that f’(x) = 2(f(x))^2?
And another question I’m starting to have doubts about is the one where they gave you the graph of f(x) and asked for the graph of f’(x)… or the opposite. was the answer the graph with a hole and jump discontinuity? or did they give the graph of f’(x) and ask for f(x)?
@Eggyolk I said m=1 and b=1 but idk if that is right. I got 4/9. And for that graph one I said the one with a jump discontinuity that had one open hole and one filled in hole. I’m not sure if that was right though.
@RHSclassof16 how did you answer that one?
Was the answer for the integral question I asked about choice A ?
The graph they provided was f(X) correct? if they did provide the graph of f(x) then I think it should be a jump discontinuity with both open-holed by f’(x) shouldn’t exist at that point because of the sharp edge.
That feeling of doubt after you finish a test…
@baller55 I don’t think that FRQ was even possible, I was thinking of writing a report for that question.
OMG DUDE!!! U ARE RIGHT BUT THAT EXAMPLE YOU USED HAS AN INTERVAL THAT CONTAINS 0!!! THE PROBLEM ON THE AP EXAM WAS AN OPEN INTERVAL AND DID NOT CONTAIN 0!!! UR EXAMPLE DOESN’T APPLY TO THE PROBLEM ON THE AP OMG
@Javacash01 no need for the all the caps lol… do you know anything about my previous post?
also for my previous post… I’m talking about Form O…
I also got 4/9. I’m sure it’s right. You had to do a U substitution and then first change the limits so they correspond with U. Then, you will notice that Du will equal something that was given in the problem to you, so substitute that in and then when you evaluate your integral i believe I did 1/2 times the integral from -1 to -1/9 of du and then I got 4/9. 4/9 is right though but i forgot about the limits of integration i used.
@Javacash01 Are we talking about the same question?
I meant the question where it gave you f(4) and f(1) or some other value and said that f’(x) = 2(f(x))^2
and asked for the value of the integral from 1 to 4 of (f(x))^2
Wait… yeah it was equal to (1/2)* f’(x) i think i got that one right i think…was that answer choice A
because i don’t remember the exact value but it was a number divided by 9 lol…
@Eggyolk I want to say 16/9 does that sound right
@Eggyolk So it was both open circles?
Also, for that question you’re talking about, choice A was 4/9 (I have the exact question + choices if you need it)
@nhljohn871 I don’t think so… I only remember the letter that I bubbled in…
@stoopidfoose ok then I got it right 
if the graph they provided was f(x) then answer is supposed to have both open circles because the derivative of a function does not exist at a “sharp” edge. So if it is 1 open circle and 1 filled in, you’re saying that the derivative exists at that point.
Also, derivative don’t exist at cusps, sharp edges like if it is a piece wise function, and at discontinuities.
@RHSclassof16 how did you do the FRQ… the one with y = mx + b?
@Javacash01 Then consider f(x) = (x - 2)^3. Now, f’(x) = 0 when x = 2 and it fulfills the condition that for 0<a<b<5, f(a)<f(b).
@Eggyolk To find the line that is a solution to the differential equation, you need to look at the second derivative. For a line, d^2y/dx^2 = 0 and the second derivative of the differential equation was d^2y/dx^2 = 2 - 2x + y. So, we set them equal to each other because we want the line to be a solution to the differential equation.
0 = 2 - 2x + y
y = 2x - 2
m = 2, b = -2.
Was d2y/dx2 really 2-2x+y?? Haha no way I actually got that but completely bs’ed it
@Eggyolk Honestly I’m not even sure. I think I just plugged in values and got them out. People who took the test at my school and people on this board got the same answers so maybe I’m right. Idk I think FRQ #4 could have historically horrible average scores.
Im really upset i was really hoping for a 5 now i think im getting a high 4. That test was so much harder then previous years. It was just everything… Does anyone else feel this way?
i know a 4 is still good but i just really wished for a 5:/. Im sad idk it just seemed a lot more time consuming and calculations for a an ap test.