🟡01 - Functions of Several Variables (Domain and Range of a function)

Рет қаралды 45,996

Күн бұрын

Пікірлер: 39

@SkanCityAcademy_SirJohn Жыл бұрын

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

@ibrahimabdulai8776 5 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?

@ibrahimabdulai8776 5 ай бұрын

I think it will only be [0,+inf) if the domain is also up to infinity

@SkanCityAcademy_SirJohn 5 ай бұрын

@ibrahimabdulai8776 it mostly depends on the function in question....

@ApochieIshmael 8 ай бұрын

Wow! indeed you are doing marvelous well in your delivery

@SkanCityAcademy_SirJohn 8 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.

@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.

@DanielOluwafemi-xd2jy 9 ай бұрын

You are certainly Ghanaian! 😂 Enjoying you from Nigeria though!

@SkanCityAcademy_SirJohn 9 ай бұрын

That's true😂😂😂 thanks so much. Guess you've had some nice experience with ghanaians

@BlessingsKanyika-v4x Ай бұрын

This channel is the best, vector functions aswell 🙌🏾🥹🙌🏾

@SkanCityAcademy_SirJohn Ай бұрын

You are most welcome

@AmenyaObedSelasi 7 ай бұрын

Please try to make short videos too please… for quick understanding for test

@SkanCityAcademy_SirJohn 7 ай бұрын

Noted with thanks

@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.

@kennedynyarko8063 Жыл бұрын

Keep it up bro

@SkanCityAcademy_SirJohn Жыл бұрын

Thank you very much

@mhlengincama9809 9 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_SirJohn 9 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....