Did uni math 39y ago (y85/86). Our professors would just write down so fast and we would copy and later teach ourselves evenings. I envy this tutor. The best there can be, simply the best!
What a blessing this video showed up in my for you page. Not only you've been able to make me understand something I've never saw in school, but the energy and the passion you put in your lesson are inspiring. The pauses to let us think and absorb the conceps before moving on make the video perfect. "Those who stop learning stop living" is now my life instructions
Sorry, I have been yelled at by my teachers so many times for not explicitly giving domain and range anytime I see a function, I now instinctively do it.
You are an excellent teacher. It makes me remember 45 years ago a teacher I had like you. It is very nice to know that there are still people with your passion and soul for teaching..👏👏👏
You explain everything so well, I wish you were my teacher when I was in school (I am getting back into doing maths, hoping to do Calculus > analysis > abstract algebra etc).
These things have been frustrating me for awhile. But today while sitting in the bathtub, it finally hit me and I understood the simplicity of the concept using t substitution. The first method was a nice touch too but for now I will simply be happy for the imoment of realization.
Beautiful approach to the question. You are too good. The mistake that most students would have make was to substitute (x + 1) into the function x*2 - 3x + 2 which is a terrible idea.
I remember back in my college days, in some books we would solve such problems by "shifting" instead of assigning a dummy variable or changing the letter; Something like: Let x → x-1 (and thus converting x+1 to x); Essentially the same but I think the terminology is somewhat less confusing than when introducing a new variable (or just a dummy letter to withhold the variable) and then assigning it back to "x"
honestly i think i prefer the direct substitution, since it better emphasizes the idea that "x" isn't special, it's just a name we're using to refer to the same number several times, and we can change it whenever we like
We can also think, that we get function g(x) = f(x + 1) = (x - 1)(x - 2) by moving f one step to the left. As we can see, roots of g are 1 and 2, so roots of f are 2 and 3 respectively. Shape of a plot won't change because of moving function one step to the left, so we get f(x) = (x - 2)(x - 3). I prefer to imagine, how function actually "looks like", before I'll dive into algebra ^^
In your second method you basically say "I will move the function back, one unit to the left". Another method is to write it in the form y=(x - p)^2 + k then add - 1 to p . Thank you for your videos. I learn from them.
You can also represent f(x) by ax^2+bx+c. Then substitute x+1 into the variable x, simplifying would give you ax^2+(2a+b)x+(a+b+c). By comparing it to f(x+1) we can find the values of a, b, and c
This concept when I did it by myself took me ages to understand, the reason was I always got confused between the the two x. In the thing is that both that x are completely different! So change one to some other letter. Then your question would make a lot of sense
I took a bit of a different take. I saw a linear become a quadratic. Thus, I can assume that the function itself is quadratic. Set f(y) = ay² + by + c. Substitute y for x+1 and solve for a, b, and c. Admittedly, the methods in this video are indeed superior.
Nice explanation… after an PhD and nearly 40 years in industry I have qualms about the way we teach “substitute” … use the “t” substitution… or use your “u” substitution… I have, more than once, had graduate engineers stumble and insist a substitution cannot be used because there already is a “t” or “u” in the equation…. Just a thought
Nice video but I'd argue that the two methods are essentially the same: the 1st is a sort of "implicit" variable substitution, the 2nd is the classical, "explicit" variable substitution we all know and love. Other than that, nicely presented as always.
A third method would be the identification. f(x) = ax^2 + bx + c f(x+1) = a(x+1)^2 + b(x+1) +c = ax^2 + 2ax + a +bx + b + c = ax^2 + (2a+b)x + a + b + c By identification: a = 1, 2a + b = -3, a + b +c = 2 b = -5, c = 6 f(×) = x^2 - 5x + 6
My method: Assume f(x) = a x^2 + b x + c. Then f(x+1) = a x^2 + (2 a + b) x +a + b + c == x^2 - 3x + 2. So a=1, 2 a + b = -3 and a + b + c = 2. We get a = 1, b = -5 and c = 6. f(x) = x^2 -5x + 6.
Could someone help me understand method 2 a bit better, I am confused and will write from what I gather; The original question asked us to find f(x) given f(x+1) How does method 2 work when we are trying to find the function f(x) not f(t)? 't' does not equal 'x' and so f(x) does not equal f(t) because of the fact that we defined 't' as t=x+1 So by saying f(t) is the solution haven't you contradicited yourself?
In this type of question, we need to know that the 'x' in the f(x+1) is not the same of the 'x' in the f(x), as that, in the method 2, the teacher renames this last one as 't'.
Very nice! 74 and still learning.
I wish to be like you.. and do maths @ age of 74..
I am 47 now..
👌Never stop learning
Because when you stop learning, you stop living 👌
I'm over eighty. This is no problem. I think I'll check out 'New Calculus' with John Gabriel now.
-See ya later!
Me too 😂😂😂
I'm only 66 and I like this training.
Am in my 60s now relearning my favorite subject in high school.
This guy was born to be a teacher; humble and yet commanding.
Did uni math 39y ago (y85/86). Our professors would just write down so fast and we would copy and later teach ourselves evenings. I envy this tutor. The best there can be, simply the best!
I did it with a slight variation of method 2: f(x) = f((x-1)+1) = (x-1)^2-3(x-1)+2 = x^2-5x+6.
My friend, you are the best math channel on YT. In fact, you are better than 99% of math professors. Thank you.
he is patient and his explanations are clear too. Make sense!!
What a blessing this video showed up in my for you page. Not only you've been able to make me understand something I've never saw in school, but the energy and the passion you put in your lesson are inspiring. The pauses to let us think and absorb the conceps before moving on make the video perfect.
"Those who stop learning stop living" is now my life instructions
Sorry, I have been yelled at by my teachers so many times for not explicitly giving domain and range anytime I see a function, I now instinctively do it.
Method 2 was so obvious once I saw it. I will never "freak out" again when I see problems of this type. Thank you!!
You are an excellent teacher. It makes me remember 45 years ago a teacher I had like you. It is very nice to know that there are still people with your passion and soul for teaching..👏👏👏
Consistently excellent work. Clear, concise and artful.
Thank you!
just replace x by (x-1) in f(x+1)=.....u got it directly
You explain everything so well, I wish you were my teacher when I was in school (I am getting back into doing maths, hoping to do Calculus > analysis > abstract algebra etc).
Wow.... me encantó su forma de mostrar lo apasionante de las matemáticas y se siente lo mucho que las disfruta... me alegró de verdad... 😊
Concept is made very clear. Love your teachings. Wish all students will make best use of teachings
These things have been frustrating me for awhile. But today while sitting in the bathtub, it finally hit me and I understood the simplicity of the concept using t substitution. The first method was a nice touch too but for now I will simply be happy for the imoment of realization.
Beautiful approach to the question. You are too good. The mistake that most students would have make was to substitute (x + 1) into the function x*2 - 3x + 2 which is a terrible idea.
Превосходно! Наконец-то вижу внятное объяснение, как решать функциональное уравнение.
This is not a functional equation, but a triviality.
I am 42 ,It’s none of my business but still trying to understand becoz in school we even don’t know the use of it great job👌👌👌
Master of teaching. That is Prime Newtons! 😊
I appreciate that you start by asking very directly "What are we trying to find?"
What a brilliant explanation sir. Loved it. Thank you.
Wow, thank God for your life. Wish I had you as my maths teacher in secondary school.
Such amath presentation is so clear and interesting ! Thanks alot sir .
Wonderful explanation!!!! Congrats!!!!
I love explaining mathematics - thanks for your efforts
Fantastic explanation!!!!
Congratulations!!!
An incredible simple class!
I remember back in my college days, in some books we would solve such problems by "shifting" instead of assigning a dummy variable or changing the letter; Something like: Let x → x-1 (and thus converting x+1 to x); Essentially the same but I think the terminology is somewhat less confusing than when introducing a new variable (or just a dummy letter to withhold the variable) and then assigning it back to "x"
honestly i think i prefer the direct substitution, since it better emphasizes the idea that "x" isn't special, it's just a name we're using to refer to the same number several times, and we can change it whenever we like
A sentence like "Let x -> x-1" gives me the heebies.
after a few tries, I came to the same method. It is easier to understand conceptually, but more prone to making arithmetic mistakes.
So I'm one of the masters😁 thank you so much for what are you doing for us in order to learn maths easily
We can also think, that we get function g(x) = f(x + 1) = (x - 1)(x - 2) by moving f one step to the left.
As we can see, roots of g are 1 and 2, so roots of f are 2 and 3 respectively. Shape of a plot won't change because of moving function one step to the left, so we get f(x) = (x - 2)(x - 3).
I prefer to imagine, how function actually "looks like", before I'll dive into algebra ^^
Tranlation with vector v=-1i
your manner of looking at the screen is really funny and you are great lecturer.
Your style is very impressive also you have command.😊
Excellent presentation. So clear.
Your are fantastic coach!
I like your way of communication!❤❤❤
I like your teaching skill. Thanks.
Parabéns pelo trabalho, acompanho seu canal pelo Brasil. Continue legendando os videos em português. ❤
In your second method you basically say "I will move the function back, one unit to the left". Another method is to write it in the form y=(x - p)^2 + k then add - 1 to p . Thank you for your videos. I learn from them.
Lovely man. Enjoyed
very beautifully explained very nice man
Simple problem but good lesson. Thank you
You can also represent f(x) by ax^2+bx+c. Then substitute x+1 into the variable x, simplifying would give you ax^2+(2a+b)x+(a+b+c). By comparing it to f(x+1) we can find the values of a, b, and c
first you'd need to prove f has to be a quadratic formula.
This concept when I did it by myself took me ages to understand, the reason was I always got confused between the the two x. In the thing is that both that x are completely different! So change one to some other letter. Then your question would make a lot of sense
I took a bit of a different take. I saw a linear become a quadratic. Thus, I can assume that the function itself is quadratic. Set f(y) = ay² + by + c. Substitute y for x+1 and solve for a, b, and c.
Admittedly, the methods in this video are indeed superior.
Amazing! Thank you
Excelente. El metodo 2. Me aclaro la razón de la necesidad del cambio de variable en integración.
Very well explained sir.
f(g(X)) =h(X). to calculate f(X) we need to calculate g-¹ supposing that g does have an inverse. So. If u= g(X) then f(u) = hog-¹(u) = h(g-¹(u)).
That was great, thanks!
f(x+1) = x² − 3x + 2
f(x) = (x − 1)² − 3(x − 1) + 2
= x² − 2x + 1 − 3x + 3 + 2
= x² − 5x + 6
Now we can find the zeroes (x-intercepts), when f(x) = 0:
x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
x₁ = 2 ∨ x₂ = 3
But that wasn't the question.
An excellent teacher
Awesome!!
Yes!!! Congratulations !!
What a nice Funda sir!!!!? Amezing.....
Very good. Thanks 🙏
Excellent teaching
Great explanation
Thanks a ton 🎉🎉
Mercie explications extraordinaire
Te felicito claro,.consiso preciso
Excellent teacher
Does the second method means we substitute the inverse function of y=x+1
It is so cool sir
Why don't we plug x-1 on the original function?It seems more intuitive than manipulating the expression to make the x+1 appear.
I personally prefer Method 1. Thanks, very well explained.
If f(x+1)=x²-3x+2 then wouldnt f(x)=(x-1)²-3(x-1)+2 and wouldnt this be a eazier way to solve this
Method 1 confuses me, method 2 I understand, thanks 👍
Maths is fun. You make it interesting.
Nice explanation… after an PhD and nearly 40 years in industry I have qualms about the way we teach “substitute” … use the “t” substitution… or use your “u” substitution… I have, more than once, had graduate engineers stumble and insist a substitution cannot be used because there already is a “t” or “u” in the equation…. Just a thought
nice very good thank u
Excellent
Excellent ,en plus le gars est très sympa !
Dammit you explain it so smoothly.
I would also go from m2 but first differentiate it then put t=x+1
It would be little bit quicker since you don't have to square
Thank you,sir
Fantastic
Very short-cut method.
Alternatively, we can replace x by (x-1) to find f(x).
Very true
Muito fera!
good..... Im 66 but continue learning still...
formidable teacher. where are you from? your English pronunciation is excellent. thank you very much.
Super❤❤❤
Whenever I see functions I freak out. But today I see light ❤❤❤.
Thanks!
Thank you. Much appreciated 👏
I repeat it once again: it cannot be explained in a clearer way. Congratulation Newtons
Nice video but I'd argue that the two methods are essentially the same: the 1st is a sort of "implicit" variable substitution, the 2nd is the classical, "explicit" variable substitution we all know and love. Other than that, nicely presented as always.
I have solved using method 2 but method 1 is very interesting.
A third method would be the identification.
f(x) = ax^2 + bx + c
f(x+1) = a(x+1)^2 + b(x+1) +c
= ax^2 + 2ax + a +bx + b + c
= ax^2 + (2a+b)x + a + b + c
By identification:
a = 1, 2a + b = -3, a + b +c = 2
b = -5, c = 6
f(×) = x^2 - 5x + 6
Cool 😎
My method:
Assume f(x) = a x^2 + b x + c.
Then f(x+1) = a x^2 + (2 a + b) x +a + b + c == x^2 - 3x + 2.
So a=1, 2 a + b = -3 and a + b + c = 2.
We get a = 1, b = -5 and c = 6.
f(x) = x^2 -5x + 6.
nice !
When mathematics became art ❤
Interesting
Could someone help me understand method 2 a bit better, I am confused and will write from what I gather;
The original question asked us to find f(x) given f(x+1)
How does method 2 work when we are trying to find the function f(x) not f(t)?
't' does not equal 'x' and so f(x) does not equal f(t) because of the fact that we defined 't' as t=x+1
So by saying f(t) is the solution haven't you contradicited yourself?
❤❤
Nice❤👍🙏🙏🙏
👏👏👏👏👏👏👏👏
f(x) = 1/2[f(x+1)+f(x-1)] -- (1)^2 ;
suppose that
f(x) = x^2 + bx + c
then f(x+1) = x^2 + (b+2)x + (b+c+1).
If f(x) = x^2 -- 3x + 2 , it means that (b+2 = --3) & (b+c+1 = 2) from which you find (b = --5) & (c = 6)
Just substitute the x in the right equation part by (x-1). That would leave you immidiately with the right solution.
In this type of question, we need to know that the 'x' in the f(x+1) is not the same of the 'x' in the f(x), as that, in the method 2, the teacher renames this last one as 't'.
شكرا
Method of Masters ✍