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..👏👏👏
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
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.
Frankly, I solved this as basically a 'word analogy' problem: x + 1 is to x as x is to x - 1. Therefore, you can solve the problem most quickly by taking the expression for f(x + 1) and if you substitute 'x - 1' in that expression everywhere you see an 'x', then simplify, you have an expression for f(x). Of course, that's is no different than doing the 't substitution' at 6:00, but it saves a step and it seems intuitively easy.
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.
Man, i just love your passion so much, you have such a good feel for teaching l. I'm literally feel safe when i watch your videos. You are an inspiration and a blessing ❤️
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 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).
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.
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
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
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 ^^
6:22 I got this same answer by thinking about it as a horizontal translation, and then shifting it back to f(x) If you're given f(x+1) (which is f(x) shifted 1 unit to the left), you can shift it back 1 unit to the right and get f(x)!
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
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.
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.
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!!
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..👏👏👏
This guy was born to be a teacher; humble and yet commanding.
Method 2 was so obvious once I saw it. I will never "freak out" again when I see problems of this type. Thank you!!
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
Consistently excellent work. Clear, concise and artful.
Thank you!
Your teaching style is super. I really congratulate you. You are great.👏
Concept is made very clear. Love your teachings. Wish all students will make best use of teachings
Wow.... me encantó su forma de mostrar lo apasionante de las matemáticas y se siente lo mucho que las disfruta... me alegró de verdad... 😊
Превосходно! Наконец-то вижу внятное объяснение, как решать функциональное уравнение.
This is not a functional equation, but a triviality.
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.
So I'm one of the masters😁 thank you so much for what are you doing for us in order to learn maths easily
Master of teaching. That is Prime Newtons! 😊
I appreciate that you start by asking very directly "What are we trying to find?"
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.
Frankly, I solved this as basically a 'word analogy' problem: x + 1 is to x as x is to x - 1. Therefore, you can solve the problem most quickly by taking the expression for f(x + 1) and if you substitute 'x - 1' in that expression everywhere you see an 'x', then simplify, you have an expression for f(x). Of course, that's is no different than doing the 't substitution' at 6:00, but it saves a step and it seems intuitively easy.
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.
Man, i just love your passion so much, you have such a good feel for teaching l.
I'm literally feel safe when i watch your videos.
You are an inspiration and a blessing ❤️
Wow, thank God for your life. Wish I had you as my maths teacher in secondary school.
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.
Обожаю африканцев, а три - это ноль и четыре тоже. Надеюсь пончо меня.
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).
Wonderful explanation!!!! Congrats!!!!
Thanks!
Thank you. Much appreciated 👏
Love the passion of the mann❤
What a brilliant explanation sir. Loved it. Thank you.
Your style is very impressive also you have command.😊
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👌👌👌
Excellent presentation. So clear.
Fantastic explanation!!!!
Congratulations!!!
thank you so much i was struggling in algebra but this has helped me tremendously
Very well explained sir.
Such amath presentation is so clear and interesting ! Thanks alot sir .
just replace x by (x-1) in f(x+1)=.....u got it directly
your manner of looking at the screen is really funny and you are great lecturer.
I like your way of communication!❤❤❤
Excellent teaching
An incredible simple class!
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.
I like your teaching skill. Thanks.
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.
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.
Your are fantastic coach!
What a nice Funda sir!!!!? Amezing.....
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.
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
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.
I love explaining mathematics - thanks for your efforts
Lovely man. Enjoyed
Parabéns pelo trabalho, acompanho seu canal pelo Brasil. Continue legendando os videos em português. ❤
An excellent teacher
very beautifully explained very nice man
Great explanation
I needed this, thank you!!! 😅
Excellent teacher
method 3 : replace directly x by x-1 --->
f(x-1+1) = f(x) = (x-1)^2 - 3(x-1) + 2 = x^2 - 2x + 1 -3x +3 +2 ---> f(x) = x^2 -5x +6
Very good. Thanks 🙏
Method 1 confuses me, method 2 I understand, thanks 👍
6:22 I got this same answer by thinking about it as a horizontal translation, and then shifting it back to f(x)
If you're given f(x+1) (which is f(x) shifted 1 unit to the left), you can shift it back 1 unit to the right and get f(x)!
I personally prefer Method 1. Thanks, very well explained.
Simple problem but good lesson. Thank you
nice very good thank u
Dammit you explain it so smoothly.
Maths is fun. You make it interesting.
Super❤❤❤
Excelente. El metodo 2. Me aclaro la razón de la necesidad del cambio de variable en integración.
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.
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)).
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
Amazing! Thank you
Amazing
Does the second method means we substitute the inverse function of y=x+1
It is so cool sir
formidable teacher. where are you from? your English pronunciation is excellent. thank you very much.
Excellent
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❤👍🙏🙏🙏
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.
Thanks a ton 🎉🎉
This channel is beautiful
Mercie explications extraordinaire
Yes!!! Congratulations !!
Fantastic
Awesome!!
Te felicito claro,.consiso preciso
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 😎
Very short-cut method.
Alternatively, we can replace x by (x-1) to find f(x).
Very true
good..... Im 66 but continue learning still...
That was great, thanks!
Thank you,sir
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.
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
When mathematics became art ❤
Solution:
f(x + 1) = x² - 3x + 2
u = x + 1 |-1
x = u - 1
f(u) = (u - 1)² - 3(u - 1) + 2
f(u) = u² - 2u + 1 - 3u + 3 + 2
f(u) = u² - 5u + 6
f(x) = x² - 5x + 6
Yes, that's what he said.
Excellent ,en plus le gars est très sympa !
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
I repeat it once again: it cannot be explained in a clearer way. Congratulation Newtons
Method of Masters ✍
Whenever I see functions I freak out. But today I see light ❤❤❤.