You are one of the best mathematics teachers l have seen ever. Your videos are too helpful for the students like us.We get immense help from your videos.These videos make our topics quite easier.Carry on.
2:45 - so... (x^2-1)/x-1 = x-1 but x cannot equal 1 for the first case but CAN be 1 for the second case of function. Wouldn't that mean that we are "wrong" when we say (x^2-1)/x-1 = x-1??? i'm confused...
Any function with a vertical asymptote must be a "Continuous" function, according to my textbooks.. My teacher told me that they will never meet, but that doesn't mean there's a gap, they'll meet at infinity.
Dave, isn't the function 1 / x - 1 not graphed as it should be? For example, if you plug x = 2, y gives back 1, and the coordinate (2, 1) is not in the curve.
hmm yeah it's not perfect, the asymptotes are correct but it's not a flawless representation. i don't really have a way to get the curves perfect as i just paste them together from a stock image of a parabola or something, please excuse the approximation!
Sorry professor, but I disagree with 2:55 . 0/0 is not undefined, it is indeterminate. Not the same. 0/0 can actually be anything we want, as any real number can be the solution to the equation "0x = 0". Therefore, since "(x-1)/(x-1)" CAN be 1 for all real values including 1, it can be safely simplified without having to exclude 1. If I'm wrong and you have the time to explain this to me, please do.
Nope, you can't do 0/0 simply because you can't divide by 0 because this implies that 0 has an inverse which is false. Because if there's an x such that 0x = 1 that's absurde because for all x, 0x = 0. so for all x, x/0 is undefined in particular, 0/0 is undefined. We talk about indeterminate when we are dealing with limits (which we were not in this case)
@@manaarikicarpentier6038 Because "0x = 1" and "0x = 0" are two completely different beasts. In the first one, "x" can't be any real number. In the second one, "x" can be any real number you want it to be.
The function is (x^2-1)/(x-1) by default hence, 1 is a point of discontinuity. But, if the function is x-1, there is no restricted domain, i.e x€R. Hope this helps you :)
because if x is equal to 1 then we get x^2 - 1/1-1 which here 1-1 makes the denominator = to 0 meaning the function will become undefined and the function would be discontinuous Hope this helped :)
Good explanation. If a function is discontinuous then why we call it A FUNCTION ANYMORE? it is in the definition of a function that its all domain members MUST have unique image But one of the domain members does not have value then Why call it a function?? If the function is producing Irrational numbers then is it a continuous or discontinuous? if a function outputs Infinity it will be a continuous or discontinuous? if a function outputs a non-terminating value it will be a continuous or discontinuous? if a function outputs indeterminate forms type values is it a continuers or discontinuous?
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Your intro and outro reminds me of some old educational cartoon I use to watch when I didn't go to school as a child. I love that vibe
You are one of the best mathematics teachers l have seen ever. Your videos are too helpful for the students like us.We get immense help from your videos.These videos make our topics quite easier.Carry on.
Your intro song was soo funny, it's like retro cartoon song
There is a lot of intensity in your voice, your words directly go into my head.😄
Great tutorial. I completely understood continuous and discontinuous with just one explanation.
Never seen a better teacher
Thank you professor Dave!
Thanks
MR PROFESSOR DAVE THANK YOU SO MUCH I DONT KNOW WHAT I WOULDVE DONE IF IT WERENT FOR YOU THANK YOU AGAIN
2:45 - so... (x^2-1)/x-1 = x-1 but x cannot equal 1 for the first case but CAN be 1 for the second case of function. Wouldn't that mean that we are "wrong" when we say (x^2-1)/x-1 = x-1??? i'm confused...
Thanks for saving me again!
bro that's about as simple as it gets. you're a fucgggging G
Thank you Sirrrr
Any function with a vertical asymptote must be a "Continuous" function, according to my textbooks..
My teacher told me that they will never meet, but that doesn't mean there's a gap, they'll meet at infinity.
Yes this videos is wrong
Sir please make a lecture on how a discontinuous function is converted into a continuous function with the use of unit step function.
Thabk you so much!!! This helps me a lot
Thank you professor dave! Liked and subbed
W you remind me of my younger days in elementary school but for harder math!
super helpful!
Very helpful
thank you professor
Thanks 👍
📚great video,keep it up!
3:05 that was totally you scratching your beard
😩
Done this lesson.
Thank you!!
Very helpful for jee aspirants 🤘
thank you!! very clear and intuitive; you do know how to teach;
i mean f(x)=5 when x
Dave, isn't the function 1 / x - 1 not graphed as it should be? For example, if you plug x = 2, y gives back 1, and the coordinate (2, 1) is not in the curve.
hmm yeah it's not perfect, the asymptotes are correct but it's not a flawless representation. i don't really have a way to get the curves perfect as i just paste them together from a stock image of a parabola or something, please excuse the approximation!
@@ProfessorDaveExplains totally excusable :) Thanks for your work!
@@ProfessorDaveExplains You should use Manim if you know Python for visualisations.
very helpful ♥
Superb sir... I understood in a easy way.. 😇
OMG THANK YOU SO MUCH GOD BLESS YOU
Your a great teacher amazing exmalantion
Well detailed explanation. Wasted 3 hr trying to understand my professor 😒
Can someone please explain me the example of piecewise function..i didn't get how the coordinates (-2,5) and (-2,-3)were found .
do you still need help
@@vodiact please explainnn 🥺
@@Butterfly-bs7jp actually i have no idea now
The way he Explains.... Like puts effort in his voice..... I get it Totally.... Humble guy 😄
Sorry professor, but I disagree with 2:55 . 0/0 is not undefined, it is indeterminate. Not the same. 0/0 can actually be anything we want, as any real number can be the solution to the equation "0x = 0". Therefore, since "(x-1)/(x-1)" CAN be 1 for all real values including 1, it can be safely simplified without having to exclude 1.
If I'm wrong and you have the time to explain this to me, please do.
Nope, you can't do 0/0 simply because you can't divide by 0 because this implies that 0 has an inverse which is false.
Because if there's an x such that 0x = 1 that's absurde because for all x, 0x = 0. so for all x, x/0 is undefined in particular, 0/0 is undefined.
We talk about indeterminate when we are dealing with limits (which we were not in this case)
@@manaarikicarpentier6038 No, sorry, you are utterly wrong about this.
@@longlostwraith5106 why?
@@manaarikicarpentier6038
Because "0x = 1" and "0x = 0" are two completely different beasts.
In the first one, "x" can't be any real number.
In the second one, "x" can be any real number you want it to be.
@@longlostwraith5106 exactly, you can't find any x such that 0x=1 so 0 isn't inversible thus you can't divide by 0
Life saver
Sir can u give me any application of discontinuity in science
Thank you Math Jesus
At 3:00 how would I know 1 is not a value if I was just seeing the simplified function of x +1??
if we put value 1 for x then it will undefined that's why here 1 cant be the value of x..
Then x + 1 is the only thing you have to be concerned about
2:17 can this function be called piecewise function?
Hello Professor Dave. At 2:54, how would i know that x cannot be equal to 1 if it shows only the simplified function?
Then you won't have to worry about it. :)
The function is (x^2-1)/(x-1) by default hence, 1 is a point of discontinuity. But, if the function is x-1, there is no restricted domain, i.e x€R. Hope this helps you :)
Then x + 1 is the only thing you have to be concerned about
because if x is equal to 1 then we get x^2 - 1/1-1 which here 1-1 makes the denominator = to 0 meaning the function will become undefined and the function would be discontinuous Hope this helped :)
I am literally Speed running mathamatics🏃💨
Bro's locking in
do you understand how the thing became -3?
so most rational functions are discontinuous functions
Good explanation. If a function is discontinuous then why we call it A FUNCTION ANYMORE? it is in the definition of a function that its all domain members MUST have unique image But one of the domain members does not have value then Why call it a function?? If the function is producing Irrational numbers then is it a continuous or discontinuous? if a function outputs Infinity it will be a continuous or discontinuous? if a function outputs a non-terminating value it will be a continuous or discontinuous? if a function outputs indeterminate forms type values is it a continuers or discontinuous?
Do u have any video that teach about limit n continuous...
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Professor Jesus Explains
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worst math crack
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