There is actually a much easier way to solve this. Since we've to find these constants c and d such that they are true for all values of x, set x = 0 to obtain an equation in c and d. Next set, x = c (or -c) and get another equation in c and d. That immediately eliminates d and leaves you with a cubic equation in c with no constant terms. That yields the results pretty quickly with some extraneous values that can be eliminated with simple substitution.
Actually, if you set x = 0 , you will get [f(c)]^2 =d and the algebra works out the same as in the video. Except that the algebra is much simpler. There is no cubic equation involved. This approach was easier for me.
@@joydeepdas4735 I think that you set x=c and x=0 maybe; if you use the undetermined value c you have to verify the solutions because you find the good c=3 and another not acceptable one.
@@annacerbara4257 Doing that means taking on trust that there is a value for c where all values of x produces a constant, but whilst the question effectively says there is, it makes me rather queasy. I prefer to prove it (and a well written question would say show that there is a constant c that produces the constant d from that expression).
I really liked your method for finding d at 10:12! When you found c using just absolute values was very cool as well. 11:40 I have never seen someone do it like that before.
I had two ways of doing it. First was to note that the expansion could be written as the the difference of two squares at the top and bottom. Hence you get ((4-c)^2-x^2)/((2c-4)^2-4x^2). Now note that we have a dividend of the form A-x^2 and a divisor of the form B-4x^2 where A & B are constants. In order for that to be invariant with the value of X, then we need to pull out a common factor 4 from the divisor so it is of the form 4(B/4-x^2). As the dividend is of the form A-x^2, it can be seen that this division will only produce a constant when B/4 = A or, re-arranged, that B = 4A. As A=(4-c^2) and B=(2c-4)^2, then (2c-4)^2 =4(4-c)^2. Expanding gives us 4c^2-16c+16=64-32C+4c^2. Simplify, and we get 16c=48, or c=3. Substitute that into the formula earlier and you find d=1/4. The other way is to differentiate the expanded expression with respect to x and you find that the gradient will be 0 when 2(c-3)=0, or c=3. Which amounts to much the same thing.
Your solution is simple and easy. Question: if expression =d is derived d'=0 and we work only on finding c. Yes is complicated and not elegant but may be in other problems it may be useful. Now a different question: Is any restrictive condition about different of 0 ? And another for c part You have (L)²=(R)² L² -R²=0 (L-R )*(L +R)=0 possible 2 solutions for c. Thank you for the examples
f(x) = -1/2 + 1/(x-2) f(c+x) f(c-x) = 1/4 + (3-c)/(something with c and x) = d To make the product invariant of x, the second term should be 0. This implies c=3, d=1/4
You can get the answer without any multiplication, by looking at where the rational function can have a zero/pole, and looking at the limit when x goes to infinity.
Hello there I love your math videos so much specially the ways you solve them makes me want to watch your videos more❤ You teach even better than my teach😂
i used the same reasoning as euclids lemma to infer that (x+c-2) divides either (x+c-4) or (x-c+4). 2nd edit: fundamental theorem of algebra does this for us.
How do we get an assumption that 1 = 4d? If it was 1 = 4d + 2 and (c - 4)² = 4d(c - 2)² + 2 the equation would be true, but c and d would be different.
How to evaluate logarithm inside a logarithm? For example log_0.8(log_144(288×3^[1÷2]))? _0.8 is the base and _144 is also the base. Using calculator the outcome is -1 but I don't want to always use a calculator for this kind of questions 😭
12:06 For those who want to see how it would look like with absolute values, here: (c-4)^2 = (c-2)^2 |c - 4| = |c - 2| Now since there are two absolute values, there's four cases to worry about: 1) +(c - 4) = +(c - 2): no solution. 2) +(c - 4) = -(c - 2): c = 3. 3) -(c - 4) = +(c - 2): c = 3 (same as #2) 4) -(c - 4) + -(c - 2): no solution (same as #1) The only way this would work is if c = 3
I made a transcription error from one line to the next and ended up deep in very complicated expressions. Do you have any advice for avoiding such errors?
Read the question, write down the first line of your working, then read it again, making sure to match your signs and values. With enough practice, this only takes a few seconds with the back-and-forth
take smaller steps. if your expressions completely changes appearance between lines, because you're doing a bunch of simplifying in your head as you go, errors are both likelier to happen and harder to notice. if you for instance do distributing in one step and combining like terms in another, as you are contemplating the _next_ step it's easy to quickly check for bugs in the _last_ step.
I tried this using variable substitutions for the two function inputs, t = c - x and u = c + x. This results in t = -u. I multiplied f(x) f(-x) and then solved for d and got d = 1/4. This is only works if c = 0 is also a solution. Am I missing something?
Once reaching (c-4)^2=(c-2)^2, I recommend solving this in the following way, a^2=b^2 => a=b or a=-b; same way c-4=c-2 leading to no solutions, and c-4=2-c giving us c=3. It's the same result but it's easier because you don't have to expand the parantheses
There is actually a much easier way to solve this. Since we've to find these constants c and d such that they are true for all values of x, set x = 0 to obtain an equation in c and d. Next set, x = c (or -c) and get another equation in c and d. That immediately eliminates d and leaves you with a cubic equation in c with no constant terms. That yields the results pretty quickly with some extraneous values that can be eliminated with simple substitution.
Actually, if you set x = 0 , you will get [f(c)]^2 =d and the algebra works out the same as in the video. Except that the algebra is much simpler. There is no cubic equation involved. This approach was easier for me.
if you place numerical values instead of x, for example x=0, x=2, verification is not necessary, which in any case is never a bad thing to do.
I did the same way and got another pair of solutions: c=0 and d=1
@@joydeepdas4735 I think that you set x=c and x=0 maybe; if you use the undetermined value c you have to verify the solutions because you find the good c=3 and another not acceptable one.
@@annacerbara4257 Doing that means taking on trust that there is a value for c where all values of x produces a constant, but whilst the question effectively says there is, it makes me rather queasy. I prefer to prove it (and a well written question would say show that there is a constant c that produces the constant d from that expression).
Great! Keep it up! I admire your lightness throughout yours solving! God bless!
I really like your exercies, it works for my to keep (like you say) learning
I really liked your method for finding d at 10:12! When you found c using just absolute values was very cool as well. 11:40 I have never seen someone do it like that before.
I had two ways of doing it. First was to note that the expansion could be written as the the difference of two squares at the top and bottom. Hence you get ((4-c)^2-x^2)/((2c-4)^2-4x^2). Now note that we have a dividend of the form A-x^2 and a divisor of the form B-4x^2 where A & B are constants. In order for that to be invariant with the value of X, then we need to pull out a common factor 4 from the divisor so it is of the form 4(B/4-x^2). As the dividend is of the form A-x^2, it can be seen that this division will only produce a constant when B/4 = A or, re-arranged, that B = 4A.
As A=(4-c^2) and B=(2c-4)^2, then (2c-4)^2 =4(4-c)^2. Expanding gives us 4c^2-16c+16=64-32C+4c^2. Simplify, and we get 16c=48, or c=3. Substitute that into the formula earlier and you find d=1/4.
The other way is to differentiate the expanded expression with respect to x and you find that the gradient will be 0 when 2(c-3)=0, or c=3. Which amounts to much the same thing.
Excelente, gracias por compartir. He comprobado. Saludos desde Chiclayo Norte del Perú
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Marvellous......
Your solution is simple and easy.
Question: if expression =d is derived
d'=0 and we work only on finding c.
Yes is complicated and not elegant but may be in other problems it may be useful.
Now a different question:
Is any restrictive condition about different of 0 ?
And another for c part
You have
(L)²=(R)²
L² -R²=0
(L-R )*(L +R)=0
possible 2 solutions for c.
Thank you for the examples
Very good. Thanks 👍
11:31 yes and thats why i like difference of two squares in this case
Thanks for the video... I guess we learn new things everyday
that was awesome!
Mind Blown
Very interesting. Thx.
f(x) = -1/2 + 1/(x-2)
f(c+x) f(c-x) = 1/4 + (3-c)/(something with c and x) = d
To make the product invariant of x, the second term should be 0.
This implies c=3, d=1/4
It seems to be difficult. Thanks for showing me the way to solve it.
You can get the answer without any multiplication, by looking at where the rational function can have a zero/pole, and looking at the limit when x goes to infinity.
Amazing
Hello Prime Newtons, you have excellent ideas of problem solving.
Glad you think so!
Good solution
You said it's easy, but you still had to do quite some work. Nothing good comes easy, my guy.
Just because something is long, doesn't mean its going to be difficult.
Dude this comment suggests that you are still immature
Hello there
I love your math videos so much specially the ways you solve them makes me want to watch your videos more❤
You teach even better than my teach😂
Thanks for such kind words
i used the same reasoning as euclids lemma to infer that (x+c-2) divides either (x+c-4) or (x-c+4). 2nd edit: fundamental theorem of algebra does this for us.
Professor can you please suggest some books which helps drastically for IMO preparation ( not SOF)
Вы молодчина.
How do we get an assumption that 1 = 4d? If it was 1 = 4d + 2 and (c - 4)² = 4d(c - 2)² + 2 the equation would be true, but c and d would be different.
That is where I got stopped as well.
Because x^2 must go away from the equation that has to be not dependent from x.
@@annacerbara4257 but why 1 = 4d, look at the example I gave.
How to evaluate logarithm inside a logarithm? For example log_0.8(log_144(288×3^[1÷2]))? _0.8 is the base and _144 is also the base. Using calculator the outcome is -1 but I don't want to always use a calculator for this kind of questions 😭
12:06 For those who want to see how it would look like with absolute values, here:
(c-4)^2 = (c-2)^2
|c - 4| = |c - 2|
Now since there are two absolute values, there's four cases to worry about:
1) +(c - 4) = +(c - 2): no solution.
2) +(c - 4) = -(c - 2): c = 3.
3) -(c - 4) = +(c - 2): c = 3 (same as #2)
4) -(c - 4) + -(c - 2): no solution (same as #1)
The only way this would work is if c = 3
going from a trivial polynomial to a multiply-piecewise abomination takes a special kind of hate :). difference of perfect squares ftw
thanks sir
Thanks for an other video....
I made a transcription error from one line to the next and ended up deep in very complicated expressions. Do you have any advice for avoiding such errors?
Read the question, write down the first line of your working, then read it again, making sure to match your signs and values. With enough practice, this only takes a few seconds with the back-and-forth
take smaller steps. if your expressions completely changes appearance between lines, because you're doing a bunch of simplifying in your head as you go, errors are both likelier to happen and harder to notice. if you for instance do distributing in one step and combining like terms in another, as you are contemplating the _next_ step it's easy to quickly check for bugs in the _last_ step.
don't need expand at bottom of video. rule is c-4 = c - 2 or c - 4 = -(c-2)
C equals to 3.
D equals to 1/4.
Really great.. Also suggestions for answers..
But isn't the solution to this based on memory, you Only remember how to do the solution??
The extreme right side is entirely invisible in the video.
I tried this using variable substitutions for the two function inputs, t = c - x and u = c + x. This results in t = -u. I multiplied f(x) f(-x) and then solved for d and got d = 1/4. This is only works if c = 0 is also a solution. Am I missing something?
2nd
Once reaching (c-4)^2=(c-2)^2, I recommend solving this in the following way, a^2=b^2 => a=b or a=-b; same way c-4=c-2 leading to no solutions, and c-4=2-c giving us c=3. It's the same result but it's easier because you don't have to expand the parantheses
عالی بود
I don't get how you found d. Where does this x^2 = 1 come from, and why can you compare those coefficients?
First!!❤
Q.) A value of x satisfying 85x ≡ 45(mod 15) is
35
10
25
15
How to solve such question sir ??
x=3k k=1,2,3 ... so the solution is x=15.
But what happens if ( C - 4 )² is not equal to ( C - 2 ) ² when you compare coefficients ?
It cannot happen if the original equation was true.
Can you please prove it, by checking your answer? Plug in the value for "c" and then FOIL {(f(3+x)with f(3-x)} and see if you get 1/4
What if f(x) male and f(c+x) *f(c-x) =d
when c is female?
Would 'd' be male or female child.
I have one for you. 2^x + 3^x = 4^x
asnwer=1.-1 isit
Meh...
Mal video muy oscuro y no se ve nada
Your Math Technique is Great Math is my favourite your Smile style in Mathematical World have specific Values PAZA M C69AoneA