FYI, you should be using the new way L'Hop needs to be written out as of 2022. Must show lim of numerator = 0 separate from lim of den = 0 before proceeding to lim of the derivatives.
there’s a mistake with part d) the area from 4 to 8 is 8-2pi but you forgot to include (even though you did the calculation) the area from 2 to 4 which is 2, so the area from 2 to 8 is 10-2pi, and if you add f(2) which is 1 that gives you that f(8) = 11-2pi
I said -3 as well for the limit (sign mistake, did arithmetic up until that + l'hopital correctly). Will I still get every point just minus the one for answer?
Hey, Mr. Tsao. Just wanted to make you aware of the proper notation for full credit on part c: lim (x->2) of 6f(x)-3x=0 lim(x->2) of x^2-5x+6=0 lim(x->2) of (6f(x)-3x)/(x^2-5x+6)-> 0/0, which is an indeterminate form. Therefore, L’Hospital’s rule can be applied. Great video though!
My teacher is in a facebook group in which leaders and creators of the AP Calculus BC exam clarify different subjects, and they have explicitly stated that this explanation should be used. I have also seen this used in the grading rubric of a recent practice exam that I took. Graders may not always be as strict as this, however.
For (d), instead of using the candidate test for absolute extrema, would it be acceptable to state the area under the f' curve is the least at x=2 and that f(2) = 1?
Yes. Since f’ is positive from (6,8), f is increasing and therefore has a relative maximum since it is a right hand endpoint. On the other hand from (-2,-1), f’ is positive so f is increasing making -2 a relative min because it is a left endpoint.
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Corrections:
Part c) I did an arithmetic mistake. The limit should be 3.
Part d) f(8) should equal 11-2pi.
Thank you! Please pin the comment.
FYI, you should be using the new way L'Hop needs to be written out as of 2022. Must show lim of numerator = 0 separate from lim of den = 0 before proceeding to lim of the derivatives.
Whoops I think I lost that point
I did not know that. That's pretty picky IMO. Where did they announce that change?
They didn't update the CED since 2020
Thanks for these early answers.
The limit for 4c should be equal to 3!
Haha I got 3 yayy
there’s a mistake with part d) the area from 4 to 8 is 8-2pi but you forgot to include (even though you did the calculation) the area from 2 to 4 which is 2, so the area from 2 to 8 is 10-2pi, and if you add f(2) which is 1 that gives you that f(8) = 11-2pi
Good catch!
I said -3 as well for the limit (sign mistake, did arithmetic up until that + l'hopital correctly). Will I still get every point just minus the one for answer?
You wont get the point for your final answer but you will for everything else
Hey, Mr. Tsao. Just wanted to make you aware of the proper notation for full credit on part c:
lim (x->2) of 6f(x)-3x=0
lim(x->2) of x^2-5x+6=0
lim(x->2) of (6f(x)-3x)/(x^2-5x+6)-> 0/0, which is an indeterminate form. Therefore, L’Hospital’s rule can be applied. Great video though!
I'll keep that in mind. Is that what actual graders use as criteria? Seems really really strict to enforce that.
My teacher is in a facebook group in which leaders and creators of the AP Calculus BC exam clarify different subjects, and they have explicitly stated that this explanation should be used. I have also seen this used in the grading rubric of a recent practice exam that I took. Graders may not always be as strict as this, however.
The denominator for frq #4 is 2(2)-5 correct? wouldnt the limit using lhopitsl be 3 using this computation?
I believe C should be positive 3 denominators of limit is 2x-5 ->2(2)-5=-1
is it okay if you didnt justify with a sign test for part a? like i just said it was neither bc the derivative doesnt switch signs
That is probably fine.
For (d), instead of using the candidate test for absolute extrema, would it be acceptable to state the area under the f' curve is the least at x=2 and that f(2) = 1?
This problem was a nice break from the rest kinda relaxing ngl 😅
For part a, what if i wrote that it's because f'(x) did not change sign at x = 6
it should be sufficient
Messed up d 😭😭but everything else good
I think you messed up the limit on part D, you changed the -5 in the denominator to -3.
Thanks for catching that. Noted in the corrections.
for part D would they take off a lot of points if I put the x value and not the y value
Not a lot I don't think.
How many points do you think d would be worth?
Part D took me forever and I didn't even get it correct. I was running out of room
should part d be for x=8, 11-2pi
You forgot the right triangle from 2 to 4
For part d, isn’t 8 a local max not local minimum ?
Yes. Since f’ is positive from (6,8), f is increasing and therefore has a relative maximum since it is a right hand endpoint.
On the other hand from (-2,-1), f’ is positive so f is increasing making -2 a relative min because it is a left endpoint.
For the minimum I put (2,3) because I thought the answer for C was the y value at 2 😭