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!
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..👏👏👏
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
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).
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 ^^
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
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
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.
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'.
This guy was born to be a teacher; humble and yet commanding.
What a perfect description of this man's instructional style. :-)
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!
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.
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!!
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..👏👏👏
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.
Consistently excellent work. Clear, concise and artful.
Thank you!
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
Master of teaching. That is Prime Newtons! 😊
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
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 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👌👌👌
What a brilliant explanation sir. Loved it. Thank you.
I appreciate that you start by asking very directly "What are we trying to find?"
Wonderful explanation!!!! Congrats!!!!
Wow, thank God for your life. Wish I had you as my maths teacher in secondary school.
Excellent teaching
Excellent presentation. So clear.
Your style is very impressive also you have command.😊
Fantastic explanation!!!!
Congratulations!!!
Превосходно! Наконец-то вижу внятное объяснение, как решать функциональное уравнение.
This is not a functional equation, but a triviality.
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
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 love explaining mathematics - thanks for your efforts
Such amath presentation is so clear and interesting ! Thanks alot sir .
An incredible simple class!
your manner of looking at the screen is really funny and you are great lecturer.
Very well explained sir.
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
I like your teaching skill. Thanks.
I like your way of communication!❤❤❤
Your are fantastic coach!
very beautifully explained very nice man
Lovely man. Enjoyed
Great explanation
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.
What a nice Funda sir!!!!? Amezing.....
Parabéns pelo trabalho, acompanho seu canal pelo Brasil. Continue legendando os videos em português. ❤
Simple problem but good lesson. Thank you
An excellent teacher
Excellent teacher
Very good. Thanks 🙏
Amazing! Thank you
nice very good thank u
just replace x by (x-1) in f(x+1)=.....u got it directly
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.
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.
Excelente. El metodo 2. Me aclaro la razón de la necesidad del cambio de variable en integración.
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)).
Yes!!! Congratulations !!
Excellent
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 personally prefer Method 1. Thanks, very well explained.
Method 1 confuses me, method 2 I understand, thanks 👍
Maths is fun. You make it interesting.
Awesome!!
good..... Im 66 but continue learning still...
It is so cool sir
Fantastic
That was great, thanks!
Te felicito claro,.consiso preciso
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.
Mercie explications extraordinaire
Excellent ,en plus le gars est très sympa !
Dammit you explain it so smoothly.
Super❤❤❤
Thanks a ton 🎉🎉
Thank you,sir
Muito fera!
I repeat it once again: it cannot be explained in a clearer way. Congratulation Newtons
Whenever I see functions I freak out. But today I see light ❤❤❤.
formidable teacher. where are you from? your English pronunciation is excellent. thank you very much.
Interesting
Nice❤👍🙏🙏🙏
When mathematics became art ❤
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 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 😎
Does the second method means we substitute the inverse function of y=x+1
nice !
Very short-cut method.
Alternatively, we can replace x by (x-1) to find f(x).
Very true
Thanks!
Thank you. Much appreciated 👏
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
شكرا
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
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.
👏👏👏👏👏👏👏👏
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
❤❤
Method of Masters ✍
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'.
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.
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.
整体换元把x+1用t代替求出 f(t)即可