I just wanted to motivate my calculus students to get 100% on their first calculus tests, so this is what I did! dear calculus students, i just got some
in-n-out gift cards for you guys ten dollars each to win one of these just make sure you get 100%
on the test this week you can do it good luck so for my calculus 1 class only one student got 100%
on the first test and for my calculus 2 class only one student got 100% on the first test
both of them will be getting the in-n-out gift cards as i promised it so congratulations to Anton
and also alex great job on the first test and now you might be wondering what we are going to
do for the second test don't worry i will tell you guys that at the end but before that i have some issues
i want to talk about this student he got the first question wrong everything else is okay
so he ended up with a 96 percent and when i saw that i was like oh my goodness oh my god
and before anything let's take a look at the first question to see how this is
so the question is asking us to find all the vertical asymptotes for the function 2x minus 6
over x squared minus 2x minus 3 and here we have the options well as you can see right here the
denominator can be factorated yeah we will just get x minus 3 times x plus 1 right yeah
but do not just set the button to be zero because remember for the vertical asymptote
we have to make sure in this kind of situation we want to get a non-zero number on the top
over zero all right and the truth is if you look at the top 2x minus 6 we can factor out
a 2 and then we get x minus 3. so in fact if you just say x is equal to 3 no that's going
to give you a 0 over 0 case that's actually a removable that removable discontinuity at x
equals three so this is not what we want all right you actually have to cancel this and you just have
to consider that x plus one you make it equal to zero so in fact x is equal to negative one so
only vertical asymptote is at x equal to negative one so the answer is b just that
um yeah unfortunately he missed it but i will tell you guys what i will do for him later on
okay so this question is question number seven on my calculus test
well we are given this limit and we are trying to see that this limit represents what and of
course this is actually the definition of the derivative equation right and if you take a
look at this right here we have h is approaching zero so this right here should remind you limit as
h approaching zero of a function and if we have a plus h minus f of a divided by h like so
then this right here is just going to be f prime of a so once we know this we can just come here
and we see that a is just 125 and then of course the function is going to be the cube root
because 125 plus h right here so the function is just the cube root of x and the intended answer is
c dandio right but the truth is i accidentally make b an answer as well you can see it
i know this is bad but like yeah this is just a small mistake on my end so let's talk about
this first if we look at this by using the following definition we'll just put it as
f prime of zero being equal to the limit as x approaching zero and then we do the function
which is f of x and the minus f of zero and then over x minus zero guess what
this is going to be the same as that except for right here we use x right but here we are using
h which is actually the same thing so it's just a notation issue but yeah
b is also the answer but the original intention was c i didn't make this wrong enough
take a look at his integral here you see that he has two inverse tangent of w good
but when he did the final substitution he forgot to write down the two
and everything else is perfect everything else is perfect so what what do you think i should do
so let's take a look at the solution for this question the integral of 1 over the square root of
e to the x minus 1. this rate is actually not so bad if we use the right method which is tricks up
and if you do u equal to the bottom and the whole thing okay the square root of e to the x minus one
this right here is going to be the quickest way to finish this question so let's go ahead and just
finish that let me just go ahead and differentiate this on both sides differentiating square root
of this we get 1 over 2 square root of the inside and then of course we have to use the chandu right
multiply by d the root of the inside which is just going to be e to the x and then don't
forget the dx right here and now i will multiply the reciprocal on both sides so we get dx being
equal to that and then we have the d u so now we can take this integral to the u world we will
just have the integral one over the whole thing here is the u and then dx is that so i will see
i'll put down two and you see that this right here it's actually the same as u so we will have 2u
over e to the x and we have the u this is pretty good because we see that the u and u cancel
and now be really careful this right here is the integral we can put a 2 in the front and we have
1 over e to the x and then in the u world so can we really integrate this the answer is no because
e to the x is not divided in the u world so the way that we have to know handle this is that
take a look right here we are going to square both sides so that we can say e to the x
minus 1 is equal to u squared and then we can just add 1 on both sides so we get e to the x is equal
to u squared plus 1. and now take a look at the e to the x we can just change that to u squared
plus one so that's the idea and this right here is the inverse tangent of u so we have two times
inverse tangent of u and u is that so let's just go ahead and put that back like this
and then we just add up c and this is it and yeah we do have a 2 right here so what am i going to do
for this student and also the other student right i am going to take away one point from
him and i'm actually going to give him a eight dollar gift card to in and out and
for the cowboys student who got 96 earlier he will get a six dollar gift card during a knot
this is for the people who got 100 on my first calculus test well i got cheesecake factory gift
card this time and this time it's 20 dollars if you can get 100 on exam 2 then you get one of this
you can do it good luck and just to make this super super fun maybe not for the students but
rather for myself i'm going to add an extra multiple choice question to my calculator test
so that the total is going to be 104 points for my calculated test and 105 points for my tattoo test
they have to get exactly 100 points in order to get a gift card so they cannot just answer all
the questions and get like one or four one 105 and get a gift card no it has to be exactly 100 points
yep yep and i will keep you guys updated so yeah that's it