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!
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..👏👏👏
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
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.
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 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.
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.
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).
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.
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.
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 ❤️
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 ^^
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
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
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.
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!
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!!
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!!
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..👏👏👏
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!
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.
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.
Превосходно! Наконец-то вижу внятное объяснение, как решать функциональное уравнение.
This is not a functional equation, but a triviality.
Concept is made very clear. Love your teachings. Wish all students will make best use of teachings
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.
Wow.... me encantó su forma de mostrar lo apasionante de las matemáticas y se siente lo mucho que las disfruta... me alegró de verdad... 😊
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.
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.
Master of teaching. That is Prime Newtons! 😊
I appreciate that you start by asking very directly "What are we trying to find?"
Your teaching style is super. I really congratulate you. You are great.👏
Wow, thank God for your life. Wish I had you as my maths teacher in secondary school.
So I'm one of the masters😁 thank you so much for what are you doing for us in order to learn maths easily
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).
just replace x by (x-1) in f(x+1)=.....u got it directly
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.
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.
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 ❤️
Wonderful explanation!!!! Congrats!!!!
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👌👌👌
thank you so much i was struggling in algebra but this has helped me tremendously
What a brilliant explanation sir. Loved it. Thank you.
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 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.
Such amath presentation is so clear and interesting ! Thanks alot sir .
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
Your style is very impressive also you have command.😊
Very well explained sir.
I love explaining mathematics - thanks for your efforts
Love the passion of the mann❤
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.
your manner of looking at the screen is really funny and you are great lecturer.
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)!
Excellent presentation. So clear.
I like your teaching skill. Thanks.
What a nice Funda sir!!!!? Amezing.....
Fantastic explanation!!!!
Congratulations!!!
Excellent teaching
I personally prefer Method 1. Thanks, very well explained.
An incredible simple class!
Thanks!
Thank you. Much appreciated 👏
I like your way of communication!❤❤❤
I needed this, thank you!!! 😅
Maths is fun. You make it interesting.
Method 1 confuses me, method 2 I understand, thanks 👍
Lovely man. Enjoyed
Your are fantastic coach!
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)).
very beautifully explained very nice man
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
Simple problem but good lesson. Thank you
nice very good thank u
Excelente. El metodo 2. Me aclaro la razón de la necesidad del cambio de variable en integración.
Excellent teacher
An excellent teacher
Very good. Thanks 🙏
Parabéns pelo trabalho, acompanho seu canal pelo Brasil. Continue legendando os videos em português. ❤
good..... Im 66 but continue learning still...
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.
Great explanation
Excellent ,en plus le gars est très sympa !
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.
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.
Dammit you explain it so smoothly.
This channel is beautiful
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
When mathematics became art ❤
Whenever I see functions I freak out. But today I see light ❤❤❤.
I have solved using method 2 but method 1 is very interesting.
Amazing
f(x+1)=x²-3x+2
Replace all x with (x-1)
f(x-1+1)=(x-1)²-3(x-1)+2
f(x)=x²-2x+1-3x+3+2
f(x)=x²-5x+6✓
This is what we call CLAY MOLDING TECHNIQUE
Does the second method means we substitute the inverse function of y=x+1
I repeat it once again: it cannot be explained in a clearer way. Congratulation Newtons
Amazing! Thank you
formidable teacher. where are you from? your English pronunciation is excellent. thank you very much.
Yes!!! Congratulations !!
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
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❤👍🙏🙏🙏
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
Super❤❤❤
Excellent
Mercie explications extraordinaire
Fantastic
It is so cool sir
Method of Masters ✍
Te felicito claro,.consiso preciso
That was great, thanks!
Just substitute the x in the right equation part by (x-1). That would leave you immidiately with the right solution.