Thank you for a great video! However, I'm kind of confused about the last one? How exactly do we find the critical points? You put the denominator equal to zero and got x= 2 and x=-2 but then you said these are not valid and that the only way to get critical points is when the denominator is 0?
Good catch! 14:24 should have said "only if the NUMERATOR is zero" not the denominator. So that's where Mr. Bean sets the numerator of the derivative "x" = 0.
@@TheAlgebros Hi, I was watching the video for 5.3 and in that video to find the critical point for example problem #1 you still included the denominator and got x=0 & x=2. & x=-2 Why is that? Do we need to find the domain of the function whenever this problem show up as well? Or plug in our answer whenever we find a DNE case to make sure x work with the original function as well?
@@minhquanduong7121 Hiii , I know Im responding late but you compare it to the domain of the original function. So you plug those values into the the original function and if it exists, then you can DNE as a critical point. If it does not fit into the original function ( not derivative), then no its not a critical point !
Thanks for the videos, but I do have one question? The extremely value theorem says these functions have to be continuous on the interval [a,b], but the first three functions have holes (so not continuous). So how could we apply the Extreme Value Theorem when we have these holes/not continuous?
That's a good point. I would argue that if you have the picture (graph), you don't have to worry about continuity because you are just looking at the visual representation of what is the "lowest" our "highest" output values. When all we have is the equations, like the last few examples, that's where we need to verify the continuity.
That one is a local (relative) minimum, but's not absolute. The reason is that we can find other y-values of that function that are smaller (below) that point. Just to the left of x=1, you can see there are y-values that are less than -1. This graph actually does not have an absolute minimum at all because of the open circle at (1, -3). Hopefully that helps!
Fantastic video thank you so much!
Glad you enjoyed it!
Thank you for a great video! However, I'm kind of confused about the last one? How exactly do we find the critical points? You put the denominator equal to zero and got x= 2 and x=-2 but then you said these are not valid and that the only way to get critical points is when the denominator is 0?
Good catch! 14:24 should have said "only if the NUMERATOR is zero" not the denominator. So that's where Mr. Bean sets the numerator of the derivative "x" = 0.
@@TheAlgebros Ohhh that makes sense thank you!!
@@TheAlgebros Hi, I was watching the video for 5.3 and in that video to find the critical point for example problem #1 you still included the denominator and got x=0 & x=2. & x=-2 Why is that? Do we need to find the domain of the function whenever this problem show up as well? Or plug in our answer whenever we find a DNE case to make sure x work with the original function as well?
@@minhquanduong7121 Hiii , I know Im responding late but you compare it to the domain of the original function. So you plug those values into the the original function and if it exists, then you can DNE as a critical point.
If it does not fit into the original function ( not derivative), then no its not a critical point !
Thanks for the videos, but I do have one question?
The extremely value theorem says these functions have to be continuous on the interval [a,b], but the first three functions have holes (so not continuous).
So how could we apply the Extreme Value Theorem when we have these holes/not continuous?
That's a good point. I would argue that if you have the picture (graph), you don't have to worry about continuity because you are just looking at the visual representation of what is the "lowest" our "highest" output values. When all we have is the equations, like the last few examples, that's where we need to verify the continuity.
Cool, that makes it more clear. Thanks for all your hard work with these videos!@@TheAlgebros
Thank you for this video! At 9:09 in the second problem, is there an absolute minimum? If not, why? Thanks!
That one is a local (relative) minimum, but's not absolute. The reason is that we can find other y-values of that function that are smaller (below) that point. Just to the left of x=1, you can see there are y-values that are less than -1. This graph actually does not have an absolute minimum at all because of the open circle at (1, -3). Hopefully that helps!
@@TheAlgebros Thank you very much!
Great vid bro
No prob, bruv
i dont think i am going to pass this AP exam
A.J.
how was it
did you end up passing
Better than I expected 😅
@@kamilamohamed284 Did you passed though?