Solving A Functional System of Equations in Two Ways
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- Опубликовано: 6 дек 2021
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I just found the inverse of g(x) and used it on the first equation.
Just realized it's literally the second solution lol
*4th* method:
f(x) = ax^2 + bx + c
To find the constant of f function g(x) = 0 then you will find f(0) = 43. Then, we need to use f(1) and f(-1) to find a and b. a + b = 14, a + b = -12; a = 1, b = 13, c = 43.
I used a third method. I let f(y)=ay^2+by+c, then plugged in f(g(x)) and solved for a, b, and c, by matching x^2’s, x’s, and constant terms.
Me too!
Your method assumes the answer is a quadratic polynomial, but the two methods of the video require no assumption.
@@bigjazbo9217 I didn't have to assume f(x) was a quadratic - I knew it was a quadratic because f(g(x)) was a quadratic as well.
@@scottleung9587 Quite true. Your method is sound because you see that f(x) is quadratic and g(x) is linear. You have the insight to see all of that. A beginner might not see it.
@@bigjazbo9217 You can confirm that the answer is a quadratic polynomial by taking the derivative with respect to x twice
f(2x - 7) = 4x^2 - 2x + 1
f'(2x - 7) = 4x - 1
f''(2x - 7) = 2
a = 2x - 7
f''(a) = 2
f'(a) = 2a + c
f''(a) = a^2 + ac + d
Okay g(x) = 2x - 7 is invertible: x(g) = (g + 7)/2. Therefore, f(g) = 4((g + 7)/2)² - 2((g + 7)/2) + 1 = g² + 13g + 43.
Therefore, f(x) = x² + 13x + 43.
Since f(2x-7) = 4x^2-2.x+1
=> f(x) is of form ax^2+b.x+c
=>f(2x-7) = a.(2x-7)^2 +b.(2x-7) +c
Some algebra and comparing coefficients
4a=4 => a=1 , 28a-2b = 2 => b=13, 49a-7b+c =1 => c=43.
f(x) = x^2 + 13x + 43
cool, i usually did this too!
For method 1 . Division of polynomials or synthetic division gives :
4x^2 - 2x + 1 = (2x -7)(2x+6) + 43 = (2x-7)(2x -7 + 13) + 43 = (2x-7)^2 + 13(2x-7) + 43
then 4x^2 - 2x + 1 = f(x)^2 + 13f(x) + 43 etc...
That's cool!
I have solved this problem using first method but after seeing video I realized that second method is quite simple . Thanks sir for this problem
Welcome 👍
You made typo mistake on your 1st method results, the correct answer is F(X)=X^2+13X+43 and not F(X)=X^2+13g(x)+43
Yes.. he forgot
8:43
So the 4x² term in f(g(x)), given that g(x) has a 2x term in it, makes it reasonable to assume that f(x) is a quadratic polynomial of the form:
f(x) = (x - p)² + q
From that, applying x=g(x)=2x + 7, we get, in the end:
p=-13/2 and q=1/4
Expanding the final equation, we get:
f(x) = x² + 13x + 43
Very nice!
I used the first method this time
I appreciate showing more than one method cuz it helps me looking at the problem from other prspectives
In our country we usually use the second method when we r given to solve such kind of problem. Whatever it takes, love both methods 😍
💖
Muy interesante.
Awesome video! Thank you!
Another great video. Videos like this one are fun and help me get over my distaste for composite functions.
Glad you enjoyed it! 🥰
f(x)=ax²+bx+c
fₒg(x)=a(g(x))²+b(g(x))+c
the equation given to us;fₒg(x)=4x²-2x+1
they are identical equations.
...
this is how I solved it.
but I realized it is not better...
Many many Thanks
Error at 5'38". 13x not 13g(x)
Thank You! I was wondering why 1 method said 13x and the other method said 13g(x) At first i was like wait...is there a new concept being taught here where we write a polynomial function using other functions as the terms of the function? (🤔 does that even exist?)
He corrects himself at 849
Very good video keep up the good work
Thanks, will do!
Really good work I appreciate you🥰
Thank you so much 🤗
The second method recognises that f(2x-7) is a linear transformation of f(x)
To retrieve f(x), we reverse the transformation: 2x-7=x' thus x=(x'+7)/2.
Substitute x=(x'+7)/2 into 4x^2-2x+1
That's right! What do you think of the other method?
Method 2 is better if you have been exposed to the required techniques. Method 1 only needs simple algebra, so I prefer Method 1.
Interesante razonamiento
I like your videos very much...i wish you will have millions of subscribers one day
This is not sufficient to conclude unless you prove that g is surjective. Finding the inverse of g does everything you need
g(x) is a 1'st degree polynomial and therefore surjective.
The invertibility of g is background information you can take for granted. No one is going to contest it.
replace x by (x+7)/2 in the 1st line
f(x)
= 4(x+7)²/4 - (x+7) +1
= x²+13x+43
g(x)=2x-7, so 2x=g(x)+7, then 4x^2 =g(x)^2+14g(x)+49. We replace all this in the expression of f(g(x)) and substitute g(x) by x.
Exactly the 2 methods I thought about instantly after I see this problem.
This is, in my understanding with some insights from my teacher:
Method (1)
(2x - 7)² = 4x² - 28x + 49
→ 4x² = (2x - 7)² + 28x - 49
→ 4x² - 2x + 1 = (2x - 7)² + 26x - 48
13(2x - 7) = 26x - 91
→ 26x = 13(2x - 7) + 91
→ 26x - 48 = 13(2x - 7) + 43
fg(x) = 4x² - 2x + 1
→ f(2x - 7) = (2x - 7)² + 13(2x - 7) + 43
→ f(x) = x² + 13x + 43
Method (2)
f(x) = fgg⁻¹(x)
g⁻¹(x) = ½(x + 7)
f(x)
= 4[½(x + 7)]² - 2[½(x + 7)] + 1
= (x + 7)² - (x + 7) + 1
= x² + 14x + 49 - x - 7 + 1
= x² + 13x + 43
- So, in my opinion, method (2) is universal for these 2 conditions; given fg(x)/gf(x) and g(x), find f(x) and also, given fg(x)/gf(x) and f(x), find g(x). A little bit of guessing but not so much as method (1) for this type of question. Method (1) is.. consider it as a little concept.. we can reduce partially the guessing problems of method (1) by doing mine.
p/s: I also check my answer using 2nd method and I did it wrong at method (1) earlier. My careless..
Very nice! 😊
Indeed, if g is an invertible function, and h = f°g is a known function, where f is unknown, then you can simply conclude that f = h°g^(-1).
別解 (1) f(x) =a・x ^2+b・x+cとおくと、(2) f(2x-7)=a(4x ^2-28x+49)+2bx+49a-7b+c=4x^2-2x+1 (3)係数を比較すると、4a=4 、--28a+2b=-2 、49a-7b+c=1 、(a、b、c)=(1、13、43) (4) よって、f(x)=x^2+13x+43
日本の受験であるあるのやつですね
自分もそれを思い付きましたが、厳密に解答しようと思ったら、なぜf (x)の最高字数が２なのかって所から示さないといけないから、あまりスマートじゃない
Instead of using y, just find x in terms of g. x= (g+7)/2. Then f(x) = f((g+7)/2)= 4((g+7)/2)^2-2(g+7)/2+1= (g+7)^2 -(g+7)+1= g^2+14g+49-g-7+1=g^2+13g+43, then replace g with x. f(x)= x^2+13x+43.
I used substitution to solve this problem: xy - cos (xy) = 1, dy/dx by implicit differentiation
awesome, first one was new and logical to me wow
Great to hear!
2nd method is the easiest....Thanks for this amazing video
Yw. Thank you!
6:00
Lol, you mistakenly put that g(x) in the final line.
Yessir!
Hi & thx. for posting.
i worked it through & came up with another statement: the setting up g(x) in regard to f(g(x)) in represented form , i think, is uncorrect. According to the Chainrule, your g(x) & f(g(x)) , should be a differentiable Functions EACH!.
Here is what i did to cheque it.
[2x-7)] for x= 1/2 ( as evaluation for both sides) = [8x- 2], meaning to find the roots of polyn. x ^2 in form of : ( x-2) ( x+6) .
the result shows: f(x) = x^2 + 4x -12.Now compairing that with general formula of ax^2+ bx + c, when a=1,......you see the sign for x^1 is minus in controdiction to plus above.
I may add your procedures ( 1 & 2 ) makes sense to me as well .Perhaps they can be explained in Complex Nomber Theory. lots of luck to you.
Thank you!
The chain rule does not say way you claim it says. What the chain rule claims is that if f is differentiable, and g is differentiable, then f°g must also be differentiable, with (f°g)' = (f'°g)·g'. However, f°g and g both being differentiable does not necessarily imply f is differentiable.
I did it using the first method. 😇 And when I watched the second method i remembered you having done similarly before so i instantly recalled the method. Thanks again! 😇🙏
Nice work!
@@SyberMath 😊
Good work 👍
Thank you so much 😀
古德古德，讲的挺好
good solution sir thanks
Happy to help
Both methode are cool 😍 , great job
Thank you! 😊
@@SyberMath ❤❤❤
@@tonyhaddad1394 💖
In first method last line u wrote g(x) instead of just x
I used another method. I inverted g(x) into x(g) and and then calculated f(x(g)) with the given f(g(x)) you then get f(g) and can simply replace g with x.
You get f(x)=x^2+13x+43.
Easiest method to replace x with ((x+14)/2)
Actually I see the thumb and thought about both solutions lol just the second one I thought about f(x)=f(g(g^-1(x)) that is basically the same method just writen different
You can also write it as f(x)=(x+6)*(x+7) + 1.
So in second method we take x as g inverese
Nice.
Thanks!
f(x) = 4((x+7)/2)^2 - 2((x+7)/2)+1
easy
Exquisite
Thank you! 😊
it is 13x not 13 g{x}
Here is a fourth method:
F(2x - 7) = 4x^2 - 2x + 1
take the derivative of both sides. Use the chain rule of course. Where the derivative of 2x-7 is equal to 2
f'(2x-7)*(2) = 8x - 2
f'(2x-7) = 4x - 1
a = 2x - 7
f'(a) = 4x - 1 - 13 + 13
f'(a) = 2a + 13
Integrate both sides with respect to a
f(a) = a^2 + 13a + c
from there it is trivial to solve for the value of c
You can also take the second derivative
f'(2x - 7) = 4x - 1
f''(2x - 7)*(2) = 4
f''(2x - 7) = 2
set a = 2x - 7
f''(a) = 2
now integrate with respect to a
f'(a) = 2a + c
integrate again
f(a) = a^2 + ac + d
replace a with x
f(x) = x^2 + cx + d, where c and d are real numbers
Now you know the form that the solution will ultimately take, you just need to find the value of c and d
f(2x-7) = (2x-7)^2 + (2x-7)*c + d = 4x^2 - 2x + 1
Yossi showed how to solve this in a different comment thread
That’s what I thought of first when I saw a composite function too ^^
That is not a valid method, because it assumes f is differentiable, and it may not be.
At the end of method one, you made a little mistake in writing: you left g(x) as part of the final expression for f(x). But I think everybody understood. I used method two at the university: faster and cleaner. Thanks for your videos.
Np. Thank you!
I prefer to use the 2nd method because it is easier. 😜😜😜
There is a more simple method based on the first one. As we know, the derivative of the complex function f(g(x)) will be the following:
(f(g(x)))'=f'(g(x))*g'(x)
This equation is true for any f() and g(). So we take derivatives for f(g(x)) and for g(x). They will be:
(f(g(x)))'=8x-2;
g'(x)=2.
And according to the equation above, we find f'(g(x)) through dividing the (f(g(x)))' by g'(x). It will be:
f'(g(x))=4x-1.
And now we put 2x-7=t. So 2t=4x-14. And cause f'(g(x))=4x-1=(4x-14)+13, f'(t) will be:
f'(t)=2t+13.
And now we take an integral to find f(t), which is an antiderivative of f'(t). It will be:
f(t)=t²+13t+C,
...where C is an unknown constant, which is to be found. To find C we take the first formula of f(g(x)) and equate it to the formula above, while replacing t with (2x-7). So we get:
4x²-2x+1=(2x-7)²+13(2x-7)+C;
4x²-2x+1=4х²-28х+49+26х-91+C;
-2x+1=-2x-42+C;
1=C-42;
C=43.
And after C is successfully found, we get the accurate formula of f(t):
f(t)=t²+13t+43.
And by replacing t with x, we finally get f(x):
f(x)=x²+13x+43.
No, you are wrong. It is not true for all f and g that (f°g)' = (f'°g)·g'. It is only true if f and g are differentiable. However, f°g and g both being differentiable does not actually imply f is differentiable, and in that case, (f°g)' may not have any simple relationship with g' at all. In this case, it only happens to work because g is invertible, so you can say f' = (f°g)'°g^(-1)/[g'°g^(-1)], which implies f is differentiable.
Replacing x by (x+7)/2 we can also get the soln
SUUUPEEER ❤❤❤❤❤🧡🧡🧡🧡🧡❤❤❤❤❤👍👍👍👍👍👍👍👍👍👍👍👍
Thank you!!! 💖😊😍
2nd method strike me ist... however i put x equals ginverse x in the f[g[x]] eqn as g is invertible then it goes on....
This time I prefer the 1st method because It revive my mind and make me remember when I was a student
Nice!
I love the second method to solve this...
the first thing that came to my mind was polynomial division
Replace x by (y+7)/2
I just let f(x) =ax^2 + bx + c, plugged g(x) into that and solved, getting a = 1, b = 13 and c = 43
I worked exactly with first method before watching the video , I liked the first method because it is about algebraic manipulation which is the smartest way in my opinion
Thank you1
@@SyberMath You're welcome and don't forget about trigonometry and functional equations next videos
I find it useful to use y = g(x) and instead of trying to find f(x), find f(y) instead. Why the different variable? Well, I'm using x for the space that is the domain of g, and y for the space that is the range of g. Sure, in the current example it's the same space (presumably), the real numbers. But in the general case when you are composing functions, it might well not be. In any case, you would then end up with f(y) = y^2 + 13y + 43, and then if a person really insists, you can "free" the variable x and use it as the argument for f.
Personally, this approach keeps me from getting confused as to whether x is the variable for the domain of g or whether it's the range of g. It avoids the temptation of abusing notation. (And for anybody who does computer programming, this is a good habit to get into.)
And it would allow you to avoid the error at 6:00, when you say "x" and write "g(x)"
I would use the chain rule to solve this question.
Good thinking!
@@SyberMath thanks🙂
This channel is better than preshtalwalkers channel(mind your decision) ,sorry presh
The answer is f(x)=x²+13x+43.
The third method is via Vandermonde matrix.
In other words, we dive into dual solution to straightforwards first two.
woow 😁🙏👍
😍😁
x^2+13x+43
useless spoiler
@@benedetti9000 true 😂😂😂
Just replace x with ((1/2)x + (7/2))
Nice note. Second method looks more like a real solution, not a trying to guess coefficients.
*Also I first used the first one 😅
Thanks
5:47 Instead of writing "g(x)" after that 13 being the coefficient, isn't it supposed to be just "x"?
yes. I forgot to write it
I use the foolish method. Let u=g(x)=2x+7 therefore x=(u+7)/2. After substitution it gives f(u)=(u^2)+(13u)+43 therefore f(x)=(x^2)+(13x)+43.
I finished this equation in fifth grade
Wow!
Fx=x2+13x+43
6.10 not 13g(x) --> 13x
That's right!
x^2 + 6x + 92
Bài toán hàm hợp dễ
Vâng
You have a mistake in the final solution of the first method. You wrote g(x) instead of x.
There is a mistake at 6:00 when you wrote g(x) instead of just x....
yes
6:01 you write x square + 13 g(x) + 43 .... that's not answer ,
it should be x square + 13x + 43
That was easy because g(x) is linear.
But that’s the wrong colour x. I’m afraid I have to not give you any marks for the first method. 😂😂😂😂😂
f(x) = x² + 13x + 43
Why?
@@SyberMath
g(x) = 2x - 7
f[g(x)] = a[g(x)]² + b.g(x) + c = a(2x- 7)² + b(2x - 7) + c = 4x² - 2x + 1
Solve this
a = 1, b = 13, c = 43
f[g(x)] = [g(x)]² + 13g(x) + 43
=> f(x) = x² + 13x + 43
f(x) =x^2+13x+43
Why?
Did you prove that f is a polynomial?
No need
So that is not math.
The second method is better and it proves that f is the polynomial.
2nd method is more saxxy.
2nd method is better
Ok thanks
Please don't do that.No use.
Don't do what?
Slow down and remember that what you’re trying to do is teach. Your numerals are sloppy and sometimes difficult to read.