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 ^^
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
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.
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 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.
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
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'.
