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.
@@MathsScienceandHinduism 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).
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
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
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😂
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.
The ratio of 2 polynomials is a constant. This means you can divide the top and bottom polynomial without remainder or formulated otherwise you can factor out the bottom.
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.
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 😭
Hi PrimeNewtons. I think it is pretty obvious that is 2 polynomials divided result in a nimber then the ratios of all coefficients will be equal and also equal to the given number. I woud say there is no need to multiply and you can just write 1/4=(c+4)^2/(4(c+2)^2)=d eo immediately d=1/4 and the squares should also be equal to eachother. Since there is no term in X we won't have an extra equation to verify coefficients
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
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?
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.
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.
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.
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
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.
