I had to make this video

[@davidmelville5675]

I tell you what, "Never stop learning. The train is coming." has an awfully ominous feel. XD

[@TimothyLoftin-l1i]

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.

[@F.E.Terman]

With all those car ads I am constantly reminded of the meaning of _range_ : it's where you can _get_ to. 😂

[@TYMCCK]

i read that as domain😂😂😂

[@Gagagag-d1w]

Yes, which is why I also got it wrong 😂

[@robertpearce8394]

That is why I prefer the terms domain and codomain.

[@angelmachuca4696]

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! 😂😂😂

[@KhalidSaifullah-te7dx]

Me also 😂😂😂

[@Sns1324]

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.

[@compiling]

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.

[@pijanV2]

When Prime Newtons uploads a community post, we all know that we are screwed

[@alipourzand6499]

Learned something on a saturday morning! I will continue living! ☺

[@emilie375]

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.

[@johnfeild6950]

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.

[@Dark々Samurai]

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.

[@mshonle]

The real train that was coming was the friends we made along the way!

[@Maths__phyics]

İ'm from Azerbaijan, İ always watch your spectacular maths video, thanks for extremely nice video

[@DadicekCz]

I think most people didn't properly read it and typed the domain instead of range

[@vara1499]

I watch your video for the problems and yout smile.

[@Kakarot-404error]

Sir pls solve it this one ❤️ K+k+k+k+k+........+k {n times} = k*k*k*k*..........*k { n times} k,n€ positive integers

[@VarunPatwal]

damm i thought you were going to use derivative,nice approach anyways!

[@RaidenTh]

I thought it was domain

[@moeberry8226]

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.

[@DevadharshanKKE]

As a jee aspirant this question is easy we use AM is greater than or equal to GM AM - arithmetic mean GM - geometric mean

[@alexandrezeddam7817]

Nice catch, though be careful with using AM-GM with negatives!

[@Anmol_Sinha]

Damn, that is a cool technique! i just differentiated to get the points where slope=0 and drew the rough graph from it 😅

[@giorgiobarchiesi5003]

Those who stop learning, will miss the train 😄 Very nice video, as always, so instructive and well presented. 😍

[@daddyerwin5097]

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

[@auroraaustralis7]

ngl i learned how to do this by finding turning points but this is actually brilliant

[@GreenMeansGOF]

I solved it with calculus

[@wernerviehhauser94]

know your technical terms and read properly.....

[@holyshit922]

But to use inverse function must be bijection

[@American_Language]

you should ask the community more questions. :]

[@PrimeNewtons]

Not really easy to come up with short questions that are interesting.

[@American_Language]

@PrimeNewtons Yeah, fair, but still! You should try asking them questions when you get the chance. It'd be pretty interesting :)

[@NadiehFan]

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.

[@redroach401]

Brcause I thought it said domain

[@SlimThrull]

I got it wrong because I misread the question. My bad.

[@arassemiaktas8075]

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.

[@rahulbamel]

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

[@kunaaaalllll]

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]

Assume angle A = angle B = angle C

[@kunaaaalllll]

@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]

@@kunaaaalllll maths is not a subject of only theory and equations it can be also related by substitution and intuition ....

[@kunaaaalllll]

@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?

[@cret859]

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.

[@emilie375]

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).

[@nedmerrill5705]

Okay, but can't you include complex numbers in the domain of f^-1 ? (Did the original question restrict the answer to be real?)

[@LordWarWar-present]

It's very interesting and unintuitive. I automatically found wrong answer (domain) and was wondered why it's not correct. Thank you!

[@dieuwer5370]

So, "domain" is about the x, "range" is about the "y"?

[@Malenbolai]

Ja

[@sakshamsharma4367]

CAN I DO THIS TYPE OF QUESTIONS BY TAKING:- ARTHMIT MEAN >= GEOMETRIC MEAN.

[@sakshamsharma4367]

this will give half part of answer, which can be use to eliminate options, i think!!

[@ashish_45playz42]

yes, that's how I did it

[@saksham3524]

.

[@robertveith6383]

Stop yelling your post in all caps.

[@Edsonrsmtm]

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.

[@ddanis6434]

i thought that you asked what was the range of x, not y 😭😭😭

[@jennymarx9228]

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

[@AbdulazizAbdurezak]

Wo amazing

[@ISuperI]

i read it as domain my brain didn't brain that time xd

[@arkadeusz91]

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 😉

[@Catalyst_3]

AM greater then equal to GM

[@adw1z]

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

[@Cen_t1369]

I did not assume x to be real

[@MogesMamo-wl4lp]

First of all the inverse is not a function! This is a bold mistake. U should graph and find turning points.

[@natel42069]

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

[@robertveith6383]

Spell out the word "You!"

[@seankeannudandan2146]

Second

[@octs609]

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.

[@GalaxyHomeA9]

Yeah i chose the first option

[@holyshit922]

Teach how to read with understanding These answers might be from a foreigners with little understanding of English

[@calculusandmathematicslearning]

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,∞)

[@JimWolfie]

In my defense i didn't fully read the question

[@Tim4141-v7u]

hahahah

[@xristos.l4259]

First

[@sagarsidhu2814]

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

[@marchrushart6594]

I am sorry, i have disappointed you. 😔

[@Metal_dead]

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

[@Malenbolai]

i THOUGHT IT WAS THE DOMAIN I'M DUMB

[@robertveith6383]

Stop yelling your post in all caps.

[@obeyy0urmaster]

Also I confused the range and domain hahahahahahaha such a horrible actuarial science professional !!!!

[@Ijkbeauty]

because (x-1)^2≥0