One of the few questions that I knew how to solve before watching your solution! And the reason I knew how to solve this type of equation is bc one time I struggled so badly when I was tutoring someone in precalc back in my college days! I would never forget about this type of equations! : )
Hey @blackpenredpen, good to see you here! 😊 We all had experiences of this type when we were faced with a problem and did not know how to solve it. You feel like you want to disappear right then and right there! 😁
@@SyberMath I don't see why anyone ANYONE would ever substitute the whole expression .why not substitute just the radical part and go from there?? That must work too..
What was that thing called Euler's substitution ? I am a class 10 student ... I solved it in a similar way but did not know that the method is specific for such questions ... but yea , I solved it ... I am happy
The first idea that comes to my mind is exactly this substitution because it is a common way to solve the integrals involving this denominator. Keep up your good work!
The lecturer seems to avoid explaining several conditions like t>0. I think that these questions probably aim to provide how to streamline complicated functions generally.
I once saw in another RUclips video that if the functional equation is of the form f(g(x))=h(x), then the solution is h(g^-1(x)), which is easy to proof. Essentially that what you did on the video, but to have it written out as an identity helps more than what one may think.
Ok, so first we set t = something, but after we determined the function f(t) the t is just a dummy variable, which gets replaced in the definition of the function when you plug in a value for t. Thus, you can replace this dummy variable "t" of the function f with anything. E.g. f(heinz) = heinz^2-1... Similar you can just replace it with an arbitrary x... In my opinion, you do not have to replace the "t" anyway, as it is just a dummy variable
I didn't understand that, and the answer of the person here didn't help me. Of course it is a f(t) "dummy variable", but that *t* in that problem is meant to have a different value than *x,* so I see no reasonable justification to switch variables like that.
@@paquirriwifi6812 I agree, it is rather missleading to just switch variables t and x, when t was dependent on x, especially as if you assign some value to t, you do not assign the same value to x. But, as long as you keep ij mind, that the variable t was just a helper variable, and the x from the start and the x in the final f(x) are DIFFERENT x, i do not see, why you shouldn't be allowed to switch.
@@georg1783 I understood, but, on my point of view, the problem was not solved, it asked *f* of the initial *x.* I imaginate solving my problems switching variables and it seems funny lol.
@@paquirriwifi6812 The Problem is solved correctly... The initial x is just some variable and it should holf for all x, so it dies not matter what you think is the "initial" x. There is no initial x. That x is just a placeholder. The proptery of the function does not change, just because you changed x to t or t to x. Or asked differently, what if my initial x = 1, yeah sure, my t would be different than 1, but it doesn't matter, because it holds for all x. ;)
Put x= sinh(t). Then you will get f(e^t)= sinht /(sinht +1). On simplification you will get f(e^t) =(e^t - e^-t) /(e^t - e^-t +2). Replacing e^t by x, you get f(x).
We can solve by using trigonometric substitution. Consider a triangle with side lengths equal to 1,x and sqrt(x^2 +1). So the function becomes f(tan(t)+sec(t))=sin(t)/(sin(t)+cos(t)) for the angle t. Then, f((sin(t)+1)/cos(t))=sin(t)/(sin(t)+cos(t))...
I have one question. Original equation gives you information about function behavior for arguments greater than 0. Yet your solution is valid for all x. Shouldnt it be indicated in some kind?
find it interesting since, you "just" have to do substitution like f(g(x)) = h(x) f(z)=h(g^-1(z)) But in real, inverses are not always well defined, so the conversion to explicit might be less "sophisticated" than the implicit definition (anything that is equivalent) 🤔 Btw x/(x+1) = 1 - 1/x. Easier to simplify
@@josiasmamani They’re just dummy variables. You can plug whatever you want into a function so long as there are no domain issues, and it is often clearer when doing that to give a different name to it, but at the end of the day, the name doesn’t matter.
@@cpotisch yes, I've just understood it, I had to see the video 3 times 😂. The problem was that he arrived to a funtion related to t, which is f(t), and in the last part he only changed that using x. (Sorry for my bad english 😅)
The problem is not totally finish : founding a function means also finding where this function is defined not just its expression. For example when you divide by 2t you got to be sure that 2t is not equal to 0.
my Solution: let f(x)=x+√(x²+1) f^-1(x) [ inverse function of f(x)]=(x²-1)/2x so f(x) =(f^-1(x)/(f^-1(x)+1) after doing algebra and simplifying f(x)=(x²-1)/(x²+2x-1)
Apparently, it might not really matter. The aim was to simplify the equation in the parameter. The first substitution was a substitution of equations, the second substitution is a substitution of labels or names. I think he should have just left it has f(t) or just called t, x0 and have an equation of f(x0).
We can replace any variable with another one. The point here is those x's do not represent the same thing anymore. So like if f(t)=2t+3 for all real t's then f(x)=2x+3 for all real x's. In this case reals is just the domain. It could be another set, too. And you are not stupid! 😊
wait what I'm confused. How can you just factor in the x like that, isn't t = x+ something something. Are we defining the x here different from the x from the one before. pls i need help
My only concern is about squaring both sides. By the way... oh God, iPad with Notability is an amazing way of studying and making math content! Nice video!!!
@@lounesz.5156 No, you can square without probleme It s only if you take the root that a ± will come a=b => a²=b² Juste multiply both side by a, a²=ab=b² cause a=b You don t need the |...|, if it's inside a square root, it's positive
When does the need to do this arise - what’s the motivation? I see that the new function is cleaner and easier to work with, but how did the original form come about - examples pls.
In real life ? There's probably no use for that Just like a lot of stuff in math, still, that's cool And being able to resolve such things also make you able to tackle more complex problemes, problemes that could be usefull in real life
@@skad2058 perhaps if you have some data-set that arrives in this form (maybe as the output of another function?) and you notice it has a simple functional relationship you can then rearrange to find the simplest form for f(x). I was hoping someone would say they’ve had to use this in the wild.
@@SingaporeSkaterSam In don't know, but I m not in the wild yet, still studing math and physics There's a lot a stuff that looks useless in real life in math, but those things could be used to demonstrate actual usefull things But those kind of equation, I've never seen those outside of the chapter on them Maybe there could be some application in real, I don t see how it could appear but I've probably no idea of what could do an enginer of his time But the process of "let t=g(x)" is verry usefull, for integral for exemple
You can use any variable you want as long as you do the replacements on both sides. I should've made this more clear. The x at the end is not the same as the x that we started with
It s not really the same t, you can replace t with x because the letter doesn't mater, like, you can write f(a)=3a or f(s)=3s if you want, it's the same thing
The only difference between the 2 definitions of f is that they have differing domains. For example, the first definition cannot be used to evaluate f(0) or f(-1) (or any negative argument for f) because you won't be able to find a value of x such that x + √(x^2+1) evaluates to 0 or -1. In fact x + √(x^2+1) is positive for all real x because the radical symbol (√) signifies a positive root. Thus, the original definition can be used to evaluate f only for a positive argument. The second definition, on the other hand , can be used to evaluate f at any argument. For example, the second definition allows you to find f(0) or f(-1). Other than this difference, the 2 definitions are identical.
min(e^x, e^2 +2-x)={e^x, x2 min(e^2 +2-x, 8)={8, x2 This means that the graph is e^x on (-infinity, 2], and e^2 +2-x on [2, infinity), and the maximum value of the function is e^2, since e^2 =~7.39
We can use any variable we want. The x and the t are generic variables not necessarily particular values. For example if f(x)=x+3 then f(t)=t+3 or f(w)=w+3 Say f(2x)=x+5 then set t=2x, we get x=t/2 f(2*t/2)=t/2+5 f(t)=t/2+5 or (t+10)/2 This means f can be written as f(x)=x/2+5 or (x+10)/2 Now let's find f(8), you can replace x with 4 in f(2x) and get f(2*4)=4+5=9 OR you can replace x with 8 in f(x) and get f(8)=(8+10)/2=9
Idk why but when I did that substitution at the beginning and I had to solve for t, my first instinct was to use the quadratic formula in reverse a = 1/2 b = -x c = 1/2 (1/2)t^2 - xt + 1/2 = 0 pretty surprising for me but considering i used reverse quotient rule for integrals and quadratic formula but solving for the constant in terms of the variable... i think my brain is headed somewhere cursed
He’s changed the input variable to something easier to work with, it ends up being called t but that’s just a label for the input (the x values on an x-y graph). He also calculates the effect on the output of changing the input in this way, which is just manipulation. Suggest you plug in a few x values to see what is going on before and after. For example an input value of x = 2 maps to the same output, 3/7, as an input of 3/4 in the original function.
There was definitely a level of rigor left out that explains why this is okay. Namely, the domain of the original function and the values that x can actually be. The replacing is 'fine' in the sense that it is shown that the same domain/range rules of the original function definition are consistent. Highly recommend to the uploader, even though some may consider this trivial, it's still incredibly important to include these key reasonings in a math presentation.
Did I? Thanks! 😊 I've always been fascinated by math! Studied math in college. Taught and tutored in math. Trained gifted students for math competitions and made problem solving a habit, a passion. I hope this answers your question
Suggestion: Find real values of k (not equal to 0) so that the equation 1/(k(x + 3)) + (6k - 3)/(x(3 - x)(x + 3)) = 1/(x(3 - x)) has 2 unique solutions which have a difference of 7. Taken from the Greek "Euclid" contest
The only thing I’ll say is that you should spend more time explaining why replacing t at the end with x is okay. The idea of a dummy variable isn’t straightforward to some students.
One of the few questions that I knew how to solve before watching your solution! And the reason I knew how to solve this type of equation is bc one time I struggled so badly when I was tutoring someone in precalc back in my college days! I would never forget about this type of equations! : )
Hey @blackpenredpen, good to see you here! 😊
We all had experiences of this type when we were faced with a problem and did not know how to solve it. You feel like you want to disappear right then and right there! 😁
@@SyberMathplease continue the series of teaching us like you did in vieta and diophantine
@@SyberMath I don't see why anyone ANYONE would ever substitute the whole expression .why not substitute just the radical part and go from there?? That must work too..
@@leif1075 except it's not
What was that thing called Euler's substitution ? I am a class 10 student ... I solved it in a similar way but did not know that the method is specific for such questions ... but yea , I solved it ... I am happy
A method that makes complicated matters easier, I also use this wonderful way.
Excellent!
The first idea that comes to my mind is exactly this substitution because it is a common way to solve the integrals involving this denominator.
Keep up your good work!
love how you tackle these challenging problems, appreciate your expertise
Glad to help!
It must be said that f is determined only for t>0.
And x différent of minus one
The lecturer seems to avoid explaining several conditions like t>0. I think that these questions probably aim to provide how to streamline complicated functions generally.
Congrats on 8k subscribers!!!
Such unique questions on here, it's incredible
Thank you! And congrats on your performance from 2 days ago
@@SyberMath Hopefully I can get to 200 subscribers soon
bro i just saw your comment on michael penn video lol
@@gatocomcirrose lmao yea I watch his videos too
Probably one of the easiest questions exploiting the nature of functions. Love your video and pls keep posting more :)
RUclips algoritması sağolsun bu güzel video ile tanışmama vesile oldu. Video için teşekkürler.
Rica ederim. Memnun oldum 😊
I once saw in another RUclips video that if the functional equation is of the form f(g(x))=h(x), then the solution is h(g^-1(x)), which is easy to proof. Essentially that what you did on the video, but to have it written out as an identity helps more than what one may think.
In 1st step t = x + √(x²+1) and x=(t² - 1)/2t
In last step t = x
How?!
Ok, so first we set t = something, but after we determined the function f(t) the t is just a dummy variable, which gets replaced in the definition of the function when you plug in a value for t. Thus, you can replace this dummy variable "t" of the function f with anything. E.g. f(heinz) = heinz^2-1... Similar you can just replace it with an arbitrary x... In my opinion, you do not have to replace the "t" anyway, as it is just a dummy variable
I didn't understand that, and the answer of the person here didn't help me. Of course it is a f(t) "dummy variable", but that *t* in that problem is meant to have a different value than *x,* so I see no reasonable justification to switch variables like that.
@@paquirriwifi6812 I agree, it is rather missleading to just switch variables t and x, when t was dependent on x, especially as if you assign some value to t, you do not assign the same value to x. But, as long as you keep ij mind, that the variable t was just a helper variable, and the x from the start and the x in the final f(x) are DIFFERENT x, i do not see, why you shouldn't be allowed to switch.
@@georg1783 I understood, but, on my point of view, the problem was not solved, it asked *f* of the initial *x.* I imaginate solving my problems switching variables and it seems funny lol.
@@paquirriwifi6812 The Problem is solved correctly... The initial x is just some variable and it should holf for all x, so it dies not matter what you think is the "initial" x. There is no initial x. That x is just a placeholder. The proptery of the function does not change, just because you changed x to t or t to x. Or asked differently, what if my initial x = 1, yeah sure, my t would be different than 1, but it doesn't matter, because it holds for all x. ;)
Substitution all the wayyy. Amazing as always !
Yes! Thank you!
Put x= sinh(t). Then you will get f(e^t)= sinht /(sinht +1). On simplification you will get f(e^t) =(e^t - e^-t) /(e^t - e^-t +2). Replacing e^t by x, you get f(x).
Nice!
An even simpler method (assuming you're familiar with hyperbolic functions):
Let x=sinh(y)
=> f(sinh(y)+(sinh(y)^2+1)^(1/2))=sinh(y)/(sinh(y)+1)
=> f(sinh(y)+cosh(y))=sinh(y)/(sinh(y)+1)
=> f(e^y)=(e^(2y)-1)/(2e^y+e^(2y)-1) . Let u=e^y
=> f(u) = (u^2-1)/(2u+u^2-1)
=> f(x) = (x^2-1)/(2x+x^2-1)
Another way to do it is by noticing that t = e^(sinh^-1(x))
So we have x = sinh( lnt )
But what you did is better
We can solve by using trigonometric substitution. Consider a triangle with side lengths equal to 1,x and sqrt(x^2 +1). So the function becomes f(tan(t)+sec(t))=sin(t)/(sin(t)+cos(t)) for the angle t. Then, f((sin(t)+1)/cos(t))=sin(t)/(sin(t)+cos(t))...
Nice!
I have one question. Original equation gives you information about function behavior for arguments greater than 0. Yet your solution is valid for all x. Shouldnt it be indicated in some kind?
Could you think of doing the t sub for the quadratic as rescaling the y axis of the graph in terms of t instead of x?
Maybe. Can I?
Super video Syber, thank you! I think I learned a nice trick with this video 😁 Please, do more videos with this kind of problems!
Will do!
@@SyberMath Thank you! Are you russian, anyway?
wow,you have widened my horizon, thanks a lot!
Glad I could help! 💖
Another great explanation, SyberMath!
Glad you think so!
Equation functions by setting implicit variables. Thanks. Is the function found satisfying?
Yes, it is
good method... nice board to explain it.. which is ?
Thank you! It is Notability
Very good video!!! By the way, what is your software to write the tutorial? Thank you!!!
Thank you and you're welcome! I use Notability with iPad.
@@SyberMath thank you for the reply!!! I´m learning a lot from your channel!!! See ya!
@@vameza1 You're welcome! This is great to hear!
At first, the problem looked easy to solve , but I couldn't solve it, but after watching the way you solved it , I realised that I was right at first.
U might not be a jee aspirant .... I solved it in class 11th ...
@@abhaypachauri7323 Yes , good for you
There must be a discussion on the domain of f(x). It will be (0, ♾️).
find it interesting since, you "just" have to do substitution like
f(g(x)) = h(x) f(z)=h(g^-1(z))
But in real, inverses are not always well defined, so the conversion to explicit might be less "sophisticated" than the implicit definition (anything that is equivalent) 🤔
Btw x/(x+1) = 1 - 1/x. Easier to simplify
I saw that if you plug in zero in the functional equation that this results into f(1)=0. A good check for the final function.
When I become a math teacher, I will give this problem to my students
Great to hear!
Wait, you can just change t to x in the final answer? But t is not equal to x? Is that allowed?
Yes, you can! That's not the same x 😁
@@SyberMath What do you mean they are not the same x? I am confused as well.
@@matthewjames6574 I'm very confused too 😥, can anyone explain it?
@@josiasmamani They’re just dummy variables. You can plug whatever you want into a function so long as there are no domain issues, and it is often clearer when doing that to give a different name to it, but at the end of the day, the name doesn’t matter.
@@cpotisch yes, I've just understood it, I had to see the video 3 times 😂. The problem was that he arrived to a funtion related to t, which is f(t), and in the last part he only changed that using x. (Sorry for my bad english 😅)
Hi bro, what's the name of the board that you use for your videos?
Notability with iPad
What equipment do you use for producing such amazing content 🔥
Thank you! iPad with pencil and Notability app 💖
@@SyberMath thank you 👍
👍
The problem is not totally finish : founding a function means also finding where this function is defined not just its expression.
For example when you divide by 2t you got to be sure that 2t is not equal to 0.
That's right!
Very good teacher 🎉❤
Thank you! 😃
参考になります！
my Solution:
let f(x)=x+√(x²+1)
f^-1(x) [ inverse function of f(x)]=(x²-1)/2x
so f(x) =(f^-1(x)/(f^-1(x)+1)
after doing algebra and simplifying f(x)=(x²-1)/(x²+2x-1)
apolgoes if this is wrong. I think this can be written as f(x)=1- cot(2arctan(x))
Sir you got quick and awesome again, substitution is the best 😊
So nice of you! 😊
How it is possible that you made t=x in the last part when t is equal to x+√x^2+1.
Apparently, it might not really matter. The aim was to simplify the equation in the parameter. The first substitution was a substitution of equations, the second substitution is a substitution of labels or names. I think he should have just left it has f(t) or just called t, x0 and have an equation of f(x0).
Thank you so much for your effort
It's my pleasure. Thank you! 💖
i have a question: why f(t) = f(x) at the end. Could tell me anyone plz, I’m stupid
We can replace any variable with another one. The point here is those x's do not represent the same thing anymore.
So like if f(t)=2t+3 for all real t's then f(x)=2x+3 for all real x's. In this case reals is just the domain. It could be another set, too.
And you are not stupid! 😊
@@SyberMath yes!
wait what I'm confused. How can you just factor in the x like that, isn't t = x+ something something. Are we defining the x here different from the x from the one before. pls i need help
no you can use x for different purposes. discard it and then get a new x
My only concern is about squaring both sides.
By the way... oh God, iPad with Notability is an amazing way of studying and making math content!
Nice video!!!
Thanks!
Yeah he missed a step, but it still works.
√(t²+1) = t - x
|t² +1| = (t-x)²
Note that ∀t∈ℝ, t²+1 > 0 so |t²+1| = t²+1
Then you have t²+1 = (t-x)²
@@lounesz.5156 No, you can square without probleme
It s only if you take the root that a ± will come
a=b => a²=b²
Juste multiply both side by a, a²=ab=b² cause a=b
You don t need the |...|, if it's inside a square root, it's positive
@@skad2058 Oh you're right ! I confused it with taking the square root of a square. Thank you!
How can I check my answer with the original problem? Just in case I'm trying to do these problems on my own I would like to be able to test my sanity.
Substitute
When does the need to do this arise - what’s the motivation? I see that the new function is cleaner and easier to work with, but how did the original form come about - examples pls.
In real life ? There's probably no use for that
Just like a lot of stuff in math, still, that's cool
And being able to resolve such things also make you able to tackle more complex problemes, problemes that could be usefull in real life
@@skad2058 perhaps if you have some data-set that arrives in this form (maybe as the output of another function?) and you notice it has a simple functional relationship you can then rearrange to find the simplest form for f(x). I was hoping someone would say they’ve had to use this in the wild.
@@SingaporeSkaterSam In don't know, but I m not in the wild yet, still studing math and physics
There's a lot a stuff that looks useless in real life in math, but those things could be used to demonstrate actual usefull things
But those kind of equation, I've never seen those outside of the chapter on them
Maybe there could be some application in real, I don t see how it could appear but I've probably no idea of what could do an enginer of his time
But the process of "let t=g(x)" is verry usefull, for integral for exemple
I like that kind of substitution.
What program are you using?
Notability
Take t,1/t,and then t-1/t=2x then substitute x in right hand side,
Yes plz make more functional eq plzzz i also want to good at these like floors
If y = √(x^2+1) + x, then 1/y = √(x^2+1) - x Therefore 2x = y - 1/y. So, f(y) = (2x) / (2x+2) = (y - 1/y) / (y - 1/y +2) = (y^2 - 1) / (y^2 -1 + 2y). *Simple* Right ?
Yes, excellent. 👌 👍
Nice!
Excellent!
Lovely ❤️
Thank you! 💖
Thanks
Why is it okay to rewrite t with x in the end of the video?
You can use any variable you want as long as you do the replacements on both sides. I should've made this more clear. The x at the end is not the same as the x that we started with
@@SyberMath ahh yes that makes more sense, thanks!
the last step you change t to x,but t not equal to x why?
you can just replace t with x at the end there?
yes!
It s not really the same t, you can replace t with x because the letter doesn't mater, like, you can write f(a)=3a or f(s)=3s if you want, it's the same thing
My question is, is this rigurous? Wouldn't you have to prove that we actually CAN set x exual to (t²-1)/(2t) ?
Under certain conditions...
😁
The only difference between the 2 definitions of f is that they have differing domains. For example, the first definition cannot be used to evaluate f(0) or f(-1) (or any negative argument for f) because you won't be able to find a value of x such that x + √(x^2+1) evaluates to 0 or -1. In fact x + √(x^2+1) is positive for all real x because the radical symbol (√) signifies a positive root. Thus, the original definition can be used to evaluate f only for a positive argument. The second definition, on the other hand , can be used to evaluate f at any argument. For example, the second definition allows you to find f(0) or f(-1). Other than this difference, the 2 definitions are identical.
@@PS-mh8ts Nice observation!
@@SyberMath Thank you.
Very good
As usual, I have another problem:
Draw the graph of
min(e^x , e^2 + 2 - x, 8 ) and hence find its maximum value .
🙏🙏🙏
min(e^x, e^2 +2-x)={e^x, x2
min(e^2 +2-x, 8)={8, x2
This means that the graph is e^x on (-infinity, 2], and e^2 +2-x on [2, infinity), and the maximum value of the function is e^2, since e^2 =~7.39
Does anyone know of a channel that teaches physics exercises on a digital whiteboard?
Check: ruclips.net/user/ThePhysicsMathsWizard
I love the way you are
Thank you! 💖
Last step was epic, that's way too hard to strike
Wow,u made it simple😲
Thank you! I didn't. Euler did! 😁
@@SyberMath ok
@@SyberMath anyway u taught me,thank u
You're welcome
Nice!
Thank you! Cheers!
this is the standardvstep what's new in it??
This problem is awesome! 🤩😁🤩
Yes for integrals this is first Euler substitution
Lovely work! Substitution and x relation to the whole God damn thing :-) how about a complex variable functions?
Nice 👏👏👏👌👌👌👍👍👍😀
Just like to note: x≠0 and x≠-1±√(2). The 0 from dividing by 2t. The other from the denominator.
from the range of x + sqrt(x^2 + 1) the domain of f is only properly defined here when x>0
Muchas gracias
De nada
Nice video!!!
Thanks!
☹️☹️anyone please explain me, can we hold x=t? And if not then why in the place of t, we are placing x
We can use any variable we want. The x and the t are generic variables not necessarily particular values.
For example if f(x)=x+3 then f(t)=t+3 or f(w)=w+3
Say f(2x)=x+5 then set t=2x, we get x=t/2
f(2*t/2)=t/2+5
f(t)=t/2+5 or (t+10)/2
This means f can be written as
f(x)=x/2+5 or (x+10)/2
Now let's find f(8), you can replace x with 4 in f(2x) and get f(2*4)=4+5=9 OR
you can replace x with 8 in f(x) and get f(8)=(8+10)/2=9
@@SyberMath oo thanks ☺sybermath, I like your videos very much. They have challenging as well as interesting problems.
@@abegbiswas3854 No problem! Thank you! 😊
Idk why but when I did that substitution at the beginning and I had to solve for t, my first instinct was to use the quadratic formula in reverse
a = 1/2
b = -x
c = 1/2
(1/2)t^2 - xt + 1/2 = 0
pretty surprising for me but considering i used reverse quotient rule for integrals and quadratic formula but solving for the constant in terms of the variable... i think my brain is headed somewhere cursed
interesting
Why does t replaced by x again?
It 's not the same x, you can write f(s)=3s or f(a)=3a, it the same, so you can write a x instead of t
@@skad2058 thanks
I have no idea why I'm watching this, but since yt recommends it to me, I guess why not🤷♂️
Maybe time to refresh math skills! 😄🤩
Thanks sir befor devide With t you must demonstrate Thatcher t#0
At the end you say: t=x. At first you say: t=x+V(x²+1). Do you mean: x=x+V(x²+1) ?
Replying so I can hear the answer as well
He’s changed the input variable to something easier to work with, it ends up being called t but that’s just a label for the input (the x values on an x-y graph). He also calculates the effect on the output of changing the input in this way, which is just manipulation. Suggest you plug in a few x values to see what is going on before and after. For example an input value of x = 2 maps to the same output, 3/7, as an input of 3/4 in the original function.
There was definitely a level of rigor left out that explains why this is okay. Namely, the domain of the original function and the values that x can actually be. The replacing is 'fine' in the sense that it is shown that the same domain/range rules of the original function definition are consistent.
Highly recommend to the uploader, even though some may consider this trivial, it's still incredibly important to include these key reasonings in a math presentation.
I agree!
x=tan(theta) may make things more juicy!
Seem hardy,but using property has solutions.
nice👍
Thanks
I pluged in x=cosha and it solved itself
Cool!
I like that exam and thank you so mach
You're welcome!
Yeah the one thing i keep forgetting
For the algorithm , ignore please
😁
One suggestion:
Do not repeat yourself.
This video could have been 50% shorter.
That's right! I do that a lot. Thanks for the feedback
RUclips algoritması ile geldim türkçe konuşan çoğu hocadan iyi anlatıyor.
Thanks for the compliment!
How did you learn math so in-depth?
Did I? Thanks! 😊
I've always been fascinated by math! Studied math in college. Taught and tutored in math. Trained gifted students for math competitions and made problem solving a habit, a passion.
I hope this answers your question
@@SyberMath why did you call this problem quick and easy?? It's not..
@@leif1075 bruh it depends on him, sometimes the question he finds difficult is easy for most of us.
@@SyberMath I thought you were into cybersecurity, actually!
@@diogenissiganos5036 I am!
I need cou de main
✋
Nice
Thanks
Sormak istediğim şey nasıl t eşittir x diyebiliyoruz
Degisken farketmez. istedigimizi kullanabiliriz ama sondaki x ile bastaki x ayni degil
Here 1-x^2/(1-x)= 1+ x = f(x) is the final answer.
Where does 1-x^2/(1-x) come from?
Suggestion:
Find real values of k (not equal to 0) so that the equation
1/(k(x + 3)) + (6k - 3)/(x(3 - x)(x + 3)) = 1/(x(3 - x))
has 2 unique solutions which have a difference of 7.
Taken from the Greek "Euclid" contest
For Indians it was very easy
🙂
Very easy
Thanks a lot 😊 (YT recommended this)
x = isin(@) works aswell
I get it
Of course it’s called EULER’s substitution.
Where is the 100k likes?
We are getting there! 🙃😂
Oh god !
Omit the range of x 🤣🤣🤣🤣
What's the range? 😂😂😂
Leave the x alone already. It doesn’t want to be found any longer. Respect its privacy!
Dear Algebra, please stop asking us to find your X. She's not coming back and don't ask Y! 😁😁😁
The only thing I’ll say is that you should spend more time explaining why replacing t at the end with x is okay. The idea of a dummy variable isn’t straightforward to some students.
Thanks for the input! I agree with you
Wow