9:00 - 10:32 The range of eg 1, ie. z = x^2 + y^2, is Z = [0, +inf) or z >=0 since any set of (x,y) will make z either 0 or a positive real number. Kindly take note.
@EdisonvsTesla Жыл бұрын
Nice
@SkanCityAcademy_SirJohn Жыл бұрын
@EdisonvsTesla thank you too
@ibrahimabdulai87764 ай бұрын
Please I don't think it is from [0,+inf) because it is from the domain that we can get the range and the domain is a closed set at the positive side as 4 which means any two values of x and y when squared, added and the result is less than or equal to 4, those are the values of x and y which will give us the range of this function. Therefore, the range will also be restricted, the maximum number that can be produced in the range is 2. Is there a situation whereby the domain is a closed set at one end and the range becomes an open set at that same end please?
@ibrahimabdulai87764 ай бұрын
I think it will only be [0,+inf) if the domain is also up to infinity
@SkanCityAcademy_SirJohn4 ай бұрын
@ibrahimabdulai8776 it mostly depends on the function in question....
@ApochieIshmael8 ай бұрын
Wow! indeed you are doing marvelous well in your delivery
@SkanCityAcademy_SirJohn8 ай бұрын
Thanks so so much...keep watching for more.
@nahomtesfaye5315 Жыл бұрын
you're good teacher thank you a lot
@SkanCityAcademy_SirJohn Жыл бұрын
You are most welcome. Thanks so much for watching as well.
@BlessingsKanyika-v4x27 күн бұрын
This channel is the best, vector functions aswell 🙌🏾🥹🙌🏾
@SkanCityAcademy_SirJohn27 күн бұрын
You are most welcome
@mathsetc. Жыл бұрын
Nice informative video. One correction suggested. The range of z=x^2+y^2 is [0, Infinity) because of the squares and not the interval as shown in the video.
@SkanCityAcademy_SirJohn Жыл бұрын
Thanks so much for the correction, I really appreciate it.
@AmenyaObedSelasi6 ай бұрын
Please try to make short videos too please… for quick understanding for test
@SkanCityAcademy_SirJohn6 ай бұрын
Noted with thanks
@DanielOluwafemi-xd2jy8 ай бұрын
You are certainly Ghanaian! 😂 Enjoying you from Nigeria though!
@SkanCityAcademy_SirJohn8 ай бұрын
That's true😂😂😂 thanks so much. Guess you've had some nice experience with ghanaians
@EdisonvsTesla Жыл бұрын
ln(y-2x)≥0=y-2x ≥1 . And y-2x>0 . I think it has two solution, we can take their intersection, thanks in advance from Ethiopia 🇪🇹
@SkanCityAcademy_SirJohn Жыл бұрын
Zero cannot be part of the domain, because Ln(0) is undefined.
@mhlengincama98098 ай бұрын
in 18:03 is'nt the range (-inf to 0)u(0 to 1) because 1 over something will always bring value less than 1?
@SkanCityAcademy_SirJohn8 ай бұрын
Yes, but then the main idea is that for 1/z + 1, which value of z will produce a define answer or value for x, and all values will do the except when z = 0. Eg. When z = 1, x = 2 z = 2, x = 1.5 and so on and on....