All teachers, if they are any good, are constantly learning. If not learning new material, they are developing different ways to explain what they know to others.
holyshit922 is correct. f(x)=x+1/x is not bijective, thus it does NOT have an inverse function! y=x/2 +/- sqrt(x^2-4)/2 is not even a function, because when x=3, y has two possible values, y=(3+sqrt(5))/2 or y=(3-sqrt(5))/2.
Same thing here XD! I read the domain and immediately reply with the first answer. Then, when I took the time to read again, I realized that it was range! 😂😂😂
In this case, we know that the function approaches infinity at 0, drops to a critical value, then rises to infinity again (and the inverse less than 0) so we can do it without finding the inverse by finding the minimum. x has a slope of 1 and 1/x has a slope of -1 at x=1 (by symmetry around the line y=x, without even calculating the derivative since it's analytic), which makes the minimum 1 + 1/1 = 2. And the maximum less than 0 is -2 by symmetry.
Well you'd have to prove that the function is odd and that the local minimum at x=1 is also the global minimum for the positive domain first, but the former is rather easy and the latter only requires some simple calculus.
I think of two ways to solve the non bijective function dilemma: To think this equations not as functions but as binary relations or to restrict domain & co-domain so it becomes bijective and can be worked on
Me, seeing the thumbnail: Well, this seems like a straightforward question, it's obviously A. So I'm obligated to watch to find out why I'm wrong. *clicks video* Me, instantly: WAIT RANGE NOT DOMAIN. *watches video anyway*
Thank you, I definitely forgot how to solve for the inverse, and it definitely felt icky back when I was learning how to do it in calc back in high school. I appreciate your video! It has shined the light on something that was not so scary, after all.
All I did was plug in values of x, observing the vertical asymptote at x=0, for positive integers the smallest output is y=2, and for negative integers the smallest output is smallest is -2, it’s easy from there.
Or AM>=GM relation for positive x to obtain x+1/x>=2 , and for negative x using same relation by multiplying with -1 on LHS(so the terms are positive) to make GM valid and then multiplying by -1 again to obtain x+1/x=
Why do you consider x = y + 1/y ?? It just clouds the explanation for no good reason. Just re-arrange your original equation as x - y + 1/x = 0. Then multiply both sides by x. That is allowable because x=0 is already disallowed by the term 1/x. You then have x^2 - yx + 1 = 0. Solve for x by the quadratic formula: x = ( y ± √(y^2 - 4) ) / 2. That is clearly the inverse function, as we have expressed x in terms of y. The discriminant must not be less than zero, so we have |y| ≥ 2. That's the condition defining the domain of the inverse function and hence the range of the original function.
I haven't dare answer because english is not my native language and I wasn't sure of the meaning of "range". Surprised to see I'm not the only one to be confused about it.
Easy: consider the _identity_ (x + 1/x)² = (x − 1/x)² + 4 Since (x − 1/x)^2 ≥ 0 for any nonzero real x, we must have (x + 1/x)² ≥ 4 for any nonzero real x and therefore | x + 1/x | ≥ 2 for any nonzero real x, meaning that the open interval (−2, 2) is excluded from the range of x + 1/x.
You made a rookie mistake bro, you have to first show that f has an inverse before solving for one variable in terms of another. A better way to solve this problem is just use first derivatives and find critical points. The derivative is 1-1/x^2 and we set it equal to 0 even though the definition of a critical point is where the derivative is 0 or undefined we don’t consider the undefined case which is x=0 since it is not in the domain of our function. After solving we obtain x=1 or -1 and plugging these values into our function we get 2 and -2 as the values.
Sir I am an Indian 15 year old student. thank you for explaining all your videos in very curious and fun way😊😊. I have very keen interest in mathematics. and through your channel I got a medium to study more about maths can you make a video on following problem? If sin A, sin B, sin C are in AP and cos A, cos B, cos C are in GP, then Value of (cos²A+Cos²C - 4 Cos A•CosC )/1 - Sin A•Sin B if video is not possible then can you please explain it in comments Thanks again
The same solution is too long to explain however I could give you a hint which is if a,b,c are in AP then 2b=a+c and if they are in GP then b²=ac also in the numerator if you observer carefully you can see it's whole squared of something 😉 - some leftover, and you'll have to use some basic formulae like sinC+sinD that's it
@DevadharshanKKE that's a particular case, you shouldn't advise students to do this or else they'll try to put values everywhere and miss the essence of mathematics
@DevadharshanKKE well you told him the easiest thing which could be asked what if the angles aren't that known like 60 40 and 80, he'd be completely clueless don't you think?
You don't know If f has a inverse. But If you clame x+1/x = r where r is real then you solve a quadratic equation for r real when delta is biger then zero.
Alternative: For x > 0: y' = 1 - 1/x^2 y'' = 2/x^3 > 0, hence y is strictly convex ==> y' = 0 is a unique global minimum here. y' = 0 ==> x^2 = 1 ==> x = 1, hence y takes minimum value 1 + 1/1 = 2. So range is y >= 2 over x < 0 (clear y->inf as x->inf). As y is odd, the range is y
Good video. I didn't participate in the survey, but I might not have been able to answer correctly. Not that I confuse the english terms 'range' or 'domain', but because of the syntax of intervals. In my native language, intervals are only noted with square brackets and their orientation indicates whether the bound is included or not. Thus, parentheses can be used to indicate the order of evaluation of expressions containing intervals. 12:34 The confusion coming from the syntactically defective expression of the interval noted (-∞,∞)⋂(-∞,-2]⋃[2,∞) is avoided in French because it would be noted ]-∞,∞[ ⋂ ( ]-∞,-2]⋃[2,∞[ ). The used one is ambiguous to me because it does not give the order of the operations ⋂ and ⋃. I expect to read ]-∞,∞[ ⋂ ( ]-∞,-2]⋃[2,∞[ ) = ]-∞,-2]⋃[2,∞[. Using a mix of parenthesis and square braquets for intervals is a mess and really confusing. The quiz notations are also difficult for me to be interpreted too, mainly because they do not respect the syntax I am used to.
En fait pour moi il manque bien des parenthèses à cet endroit (enfin j'imagine que les parenthèses qu'on voit sont celles de l'intervalle ouvert et pas de la priorité des opérations).
Does range not mean the values y can take? The video suggests that even if when we can choose the value of x as anything we want, we still could not make y any value between 2 and -2. But if I take x to be a complex number I could make the value of y between 2 and -2.
Yes, the question is ill-defined in a technical sense. It makes the implicit assumption that the domain is the “natural real domain” of the function. It should have specified the domain for the question to make sense. I could take the domain to be {1}, in which case the range is just {2}.
When i answered, I thought -infinity to -1 and 1 to infinity becoz i forgot to add those 1 and 1 at x=1... I did this plotting the graph Like x= 0.9 = 0.9+1.11= >2 X= 1.1 = 1.1 + 0.90909 =>2 But when x = 1, i tmforgot that there's two of 1s hanging out there and just thought it was x+1/x = 1 But after that green symbol, i realised 😢... Blunder
Its like when the teacher is not happy with the exam results so he comes back on the exam problem to explain to the whole class 😂😂 Thank you for the videos sir haha
another method is by AM GM INEQUALITY BUT IT ONLY ALLOWS U TO FIND THE +VE PART BUT IF U SUBSTITUTTE -VE X in that ineuality u will get the other part too
OHHHH, see I read it as (x+1)/(x) not knowing it was x+(1/x). I had assumed the horizontal asymptote was just the radio of the coefficients y=1 in that case.
All teachers, if they are any good, are constantly learning. If not learning new material, they are developing different ways to explain what they know to others.
I tell you what, "Never stop learning. The train is coming." has an awfully ominous feel. XD
holyshit922 is correct. f(x)=x+1/x is not bijective, thus it does NOT have an inverse function! y=x/2 +/- sqrt(x^2-4)/2 is not even a function, because when x=3, y has two possible values, y=(3+sqrt(5))/2 or y=(3-sqrt(5))/2.
With all those car ads I am constantly reminded of the meaning of _range_ : it's where you can _get_ to. 😂
Domain is where you start, such as a garage. Range is where you end up, such as stuck in a mud oit after the car ad lied to you.
nice vid! this really helped me understand a new way of solving for range and a question like this would have gotten me stuck on an exam :)
i read that as domain😂😂😂
Yes, which is why I also got it wrong 😂
That is why I prefer the terms domain and codomain.
Same thing here XD! I read the domain and immediately reply with the first answer. Then, when I took the time to read again, I realized that it was range! 😂😂😂
Me also 😂😂😂
Yep same
In this case, we know that the function approaches infinity at 0, drops to a critical value, then rises to infinity again (and the inverse less than 0) so we can do it without finding the inverse by finding the minimum. x has a slope of 1 and 1/x has a slope of -1 at x=1 (by symmetry around the line y=x, without even calculating the derivative since it's analytic), which makes the minimum 1 + 1/1 = 2. And the maximum less than 0 is -2 by symmetry.
Well you'd have to prove that the function is odd and that the local minimum at x=1 is also the global minimum for the positive domain first, but the former is rather easy and the latter only requires some simple calculus.
I think of two ways to solve the non bijective function dilemma:
To think this equations not as functions but as binary relations
or
to restrict domain & co-domain so it becomes bijective and can be worked on
Me, seeing the thumbnail: Well, this seems like a straightforward question, it's obviously A. So I'm obligated to watch to find out why I'm wrong.
*clicks video*
Me, instantly: WAIT RANGE NOT DOMAIN.
*watches video anyway*
They didn't get the *answer* wrong, they got the *question* wrong. I think the 60% - or at least a major part of it - mixed up 'range' with 'domain'.
Thank you, I definitely forgot how to solve for the inverse, and it definitely felt icky back when I was learning how to do it in calc back in high school. I appreciate your video! It has shined the light on something that was not so scary, after all.
All I did was plug in values of x, observing the vertical asymptote at x=0, for positive integers the smallest output is y=2, and for negative integers the smallest output is smallest is -2, it’s easy from there.
12:07 Should the two intervals in the union technically be in brackets?
Clearly it doesn’t make a difference but for the sake of completeness.
Or AM>=GM relation for positive x to obtain x+1/x>=2 , and for negative x using same relation by multiplying with -1 on LHS(so the terms are positive) to make GM valid and then multiplying by -1 again to obtain x+1/x=
(Arithmetic Mean) >or= (Geometric mean)
I have a question I got in olympiadic test
a,b,c are positive integers show that if:
(a/b)+(b/c)+(c/a) is an integer
then a*b*c is a cube
When Prime Newtons uploads a community post, we all know that we are screwed
Learned something on a saturday morning! I will continue living! ☺
Why do you consider x = y + 1/y ?? It just clouds the explanation for no good reason.
Just re-arrange your original equation as x - y + 1/x = 0. Then multiply both sides by x. That is allowable because x=0 is already disallowed by the term 1/x.
You then have x^2 - yx + 1 = 0. Solve for x by the quadratic formula: x = ( y ± √(y^2 - 4) ) / 2. That is clearly the inverse function, as we have expressed x in terms of y.
The discriminant must not be less than zero, so we have |y| ≥ 2. That's the condition defining the domain of the inverse function and hence the range of the original function.
I haven't dare answer because english is not my native language and I wasn't sure of the meaning of "range". Surprised to see I'm not the only one to be confused about it.
Easy: consider the _identity_
(x + 1/x)² = (x − 1/x)² + 4
Since (x − 1/x)^2 ≥ 0 for any nonzero real x, we must have
(x + 1/x)² ≥ 4
for any nonzero real x and therefore
| x + 1/x | ≥ 2
for any nonzero real x, meaning that the open interval (−2, 2) is excluded from the range of x + 1/x.
Question: why did you write (-∞, ∞) ∩ before the domain of f⁻¹(x) ?
Isnt that like saying plus 0 or times 1 ?
I got it right. I can never forget the answer of this question. This question has been asked in my exams quite a few times.
another way to solve it is by taking the derivative
As a jee aspirant this question is easy
we use AM is greater than or equal to GM
AM - arithmetic mean
GM - geometric mean
Nice catch, though be careful with using AM-GM with negatives!
Damn, that is a cool technique! i just differentiated to get the points where slope=0 and drew the rough graph from it 😅
Yeah, got to the part where I needed to solve the inverse for y, then I got stuck
i think you ask about (domain) and i get mad really mad , until i saw this video i realized that is was (rangeee!) 😂
I watch your video for the problems and yout smile.
I didn't knew bro was just like that 😎
The real train that was coming was the friends we made along the way!
Never heard of "range", only"image" and "domain"
Image and range are pretty much completely synonymous so it’s not that bad
answer=(b) isit
I thought it was domain
You made a rookie mistake bro, you have to first show that f has an inverse before solving for one variable in terms of another. A better way to solve this problem is just use first derivatives and find critical points. The derivative is 1-1/x^2 and we set it equal to 0 even though the definition of a critical point is where the derivative is 0 or undefined we don’t consider the undefined case which is x=0 since it is not in the domain of our function. After solving we obtain x=1 or -1 and plugging these values into our function we get 2 and -2 as the values.
damm i thought you were going to use derivative,nice approach anyways!
I solved it with calculus
I think most people didn't properly read it and typed the domain instead of range
you should ask the community more questions. :]
Not really easy to come up with short questions that are interesting.
@PrimeNewtons Yeah, fair, but still! You should try asking them questions when you get the chance. It'd be pretty interesting :)
Sir I am an Indian 15 year old student. thank you for explaining all your videos in very curious and fun way😊😊. I have very keen interest in mathematics. and through your channel I got a medium to study more about maths
can you make a video on following problem?
If sin A, sin B, sin C are in AP and cos A, cos B, cos C are in GP, then
Value of (cos²A+Cos²C - 4 Cos A•CosC )/1 - Sin A•Sin B
if video is not possible then can you please explain it in comments
Thanks again
The same solution is too long to explain however I could give you a hint which is if a,b,c are in AP then 2b=a+c and if they are in GP then b²=ac also in the numerator if you observer carefully you can see it's whole squared of something 😉 - some leftover, and you'll have to use some basic formulae like sinC+sinD that's it
Assume angle A = angle B = angle C
@DevadharshanKKE that's a particular case, you shouldn't advise students to do this or else they'll try to put values everywhere and miss the essence of mathematics
@@kunaaaalllll maths is not a subject of only theory and equations it can be also related by substitution and intuition ....
@DevadharshanKKE well you told him the easiest thing which could be asked what if the angles aren't that known like 60 40 and 80, he'd be completely clueless don't you think?
İ'm from Azerbaijan, İ always watch your spectacular maths video, thanks for extremely nice video
You don't know If f has a inverse. But If you clame x+1/x = r where r is real then you solve a quadratic equation for r real when delta is biger then zero.
Brcause I thought it said domain
Those who stop learning, will miss the train 😄
Very nice video, as always, so instructive and well presented. 😍
Alternative: For x > 0: y' = 1 - 1/x^2
y'' = 2/x^3 > 0, hence y is strictly convex ==> y' = 0 is a unique global minimum here.
y' = 0 ==> x^2 = 1 ==> x = 1, hence y takes minimum value 1 + 1/1 = 2. So range is y >= 2 over x < 0 (clear y->inf as x->inf).
As y is odd, the range is y
Good video.
I didn't participate in the survey, but I might not have been able to answer correctly. Not that I confuse the english terms 'range' or 'domain', but because of the syntax of intervals. In my native language, intervals are only noted with square brackets and their orientation indicates whether the bound is included or not. Thus, parentheses can be used to indicate the order of evaluation of expressions containing intervals.
12:34 The confusion coming from the syntactically defective expression of the interval noted (-∞,∞)⋂(-∞,-2]⋃[2,∞) is avoided in French because it would be noted ]-∞,∞[ ⋂ ( ]-∞,-2]⋃[2,∞[ ). The used one is ambiguous to me because it does not give the order of the operations ⋂ and ⋃.
I expect to read ]-∞,∞[ ⋂ ( ]-∞,-2]⋃[2,∞[ ) = ]-∞,-2]⋃[2,∞[. Using a mix of parenthesis and square braquets for intervals is a mess and really confusing.
The quiz notations are also difficult for me to be interpreted too, mainly because they do not respect the syntax I am used to.
En fait pour moi il manque bien des parenthèses à cet endroit (enfin j'imagine que les parenthèses qu'on voit sont celles de l'intervalle ouvert et pas de la priorité des opérations).
I got it wrong because I misread the question. My bad.
But to use inverse function must be bijection
Does range not mean the values y can take? The video suggests that even if when we can choose the value of x as anything we want, we still could not make y any value between 2 and -2. But if I take x to be a complex number I could make the value of y between 2 and -2.
Yes, the question is ill-defined in a technical sense. It makes the implicit assumption that the domain is the “natural real domain” of the function. It should have specified the domain for the question to make sense. I could take the domain to be {1}, in which case the range is just {2}.
Y = x + 1/x
Y = (x²+1)/x
xY= x²+1
x²-xY+1=0
Y²-4>=0. {D=b²-4ac>=0 ; for
real roots}
(Y-2)(Y+2)>=0
Y=(- ♾️, -2)union (2, ♾️)
When i answered, I thought -infinity to -1 and 1 to infinity becoz i forgot to add those 1 and 1 at x=1...
I did this plotting the graph
Like x= 0.9 = 0.9+1.11= >2
X= 1.1 = 1.1 + 0.90909 =>2
But when x = 1, i tmforgot that there's two of 1s hanging out there and just thought it was x+1/x = 1
But after that green symbol, i realised 😢... Blunder
know your technical terms and read properly.....
Its like when the teacher is not happy with the exam results so he comes back on the exam problem to explain to the whole class 😂😂
Thank you for the videos sir haha
ngl i learned how to do this by finding turning points but this is actually brilliant
It's very interesting and unintuitive. I automatically found wrong answer (domain) and was wondered why it's not correct. Thank you!
Wo amazing
CAN I DO THIS TYPE OF QUESTIONS BY TAKING:- ARTHMIT MEAN >= GEOMETRIC MEAN.
this will give half part of answer, which can be use to eliminate options, i think!!
yes, that's how I did it
.
Stop yelling your post in all caps.
First of all the inverse is not a function! This is a bold mistake. U should graph and find turning points.
Well the inverse that is shown is technically 2 functions, but both of them have the proper domain to show the range of the initial graph
Spell out the word "You!"
I never learned maths in English so I also made the same mistake of mixing range and domain 😅 but I learned the correct terminology now 😉
Okay, but can't you include complex numbers in the domain of f^-1 ?
(Did the original question restrict the answer to be real?)
Second
i read it as domain
my brain didn't brain that time xd
yx=x²+1
x²-yx+1=0
x= y±√y²-4/2
f^-1(y) = y±√y²-4/2
f^-¹(x)= x±√x²-4/2
x²-4≥0
(x+2)(x-2)≥0
-∞. -2. 2. ∞
+Ve -ve. +Ve
Dom of f^-¹(x)= (-∞,-2]U[2,∞)
another method is by AM GM INEQUALITY BUT IT ONLY ALLOWS U TO FIND THE +VE PART BUT IF U SUBSTITUTTE -VE X in that ineuality u will get the other part too
I am sorry, i have disappointed you. 😔
AM greater then equal to GM
I did not assume x to be real
OHHHH, see I read it as (x+1)/(x) not knowing it was x+(1/x). I had assumed the horizontal asymptote was just the radio of the coefficients y=1 in that case.
Teach how to read with understanding
These answers might be from a foreigners with little understanding of English
First
hahahah
i thought that you asked what was the range of x, not y 😭😭😭
I found another proof. First of all y(-x)=-y(x)
For x>0 => x+1/x=2(x+1/x)/2 >= 2*x/x=2. And when x=1 y=2. So, y>= 2 for x>0. And for simmetry y
In my defense i didn't fully read the question
Yeah i chose the first option
i THOUGHT IT WAS THE DOMAIN I'M DUMB
Stop yelling your post in all caps.
Also I confused the range and domain hahahahahahaha such a horrible actuarial science professional !!!!
because (x-1)^2≥0
I have a question I got in olympiadic test
a,b,c are positive integers show that if:
(a/b)+(b/c)+(c/a) is an integer
then a*b*c is a cube
I have a question I got in olympiadic test
a,b,c are positive integers show that if:
(a/b)+(b/c)+(c/a) is an integer
then a*b*c is a cube