Calculus problem for algebra students!

[@bprpmathbasics]

55% got this inequality wrong! kzbin.info/www/bejne/rJuqYnecbKmpbLs

[@maxgoldman8903]

Another method: max y is equivalent to min 1/y=(x²+2x+1)/x=2+x+1/x. Now the question is reduced to min x+1/x. By the AM-GM inequality (or use the fact that a²+b²≥ 2ab), x+1/x≥ 2. Hence, min 1/y = 4, which implies max y=1/4.

[@FNShadow008]

Amazing method!

[@discreaminant]

Only work for positive x btw

[@FNShadow008]

@@discreaminant True

[@FNShadow008]

@@discreaminant But I think for this particular case it's still a correct method, since if we have x0, y>0 as well, so to find the maximum value of y, we can assume y is positive (because positive values of y exist and would be greater than the negative values) and therefore, since y is positive, we can assume that x is positive as well

[@maxgoldman8903]

@@discreaminant Clearly, y ≥ 0 for x ≥ 0, so max y is attained at some positive x. This implies min (x+1/x)=2.

[@danielbranscombe6662]

another method x/(x+1)^2=1/(x+1)-1/(x+1)^2 so use substitution u=1/(x+1) y=u-u^2 y=1/4-(u-1/2)^2 so y is a parabola in terms of u, it opens downwards and has it's vertex at (1/2,1/4) Thus y achieves a max value of 1/4 when u=1/2 1/(x+1)=u 1/(x+1)=1/2 x+1=2 x=1

[@bprpmathbasics]

Oh that’s very nice! I didn’t think of that!

[@JayTemple]

That's how I did it too!

[@StephenBoothUK]

I just thought about the shape of the graph. Where X is greater than 1, (X+1)^2 is greater than 4 and the larger X gets the smaller Y gets. If 0 < X < 1 then (X+1)^2 > X so the smaller X gets the smaller Y gets. Therefore the maximum value of Y must be the value at X=1.

[@Meh-j9s]

Thank you for your videos. I took calculus courses 35 years ago and now my kiddo is in middle school so I’m brushing up. You make math fun.

[@wilfredrohlfing7738]

Max occurs for x>0, so x=a^2 giving y= 1/[ a + 1/a ]^2 = 1 / [ 4 + ( a - 1/a)^2 ]. Max is when denominator minimum. Obviously, this is 4 when a=1, giving y= 1/4 at x=1

[@Wildcard71]

(own try) First find the derivation y' = ((x+1)² - x*2(x+1))/(x+1)⁴ = ((x+1) - 2x)/(x+1)³ = (1-x)/(x+1)³ Set y'=0: 1-x=0; x=1 x=1; y=1/4

[@paulfillingham4778]

You can solve this intuitively. I just looked at the equation and it’s clear that the greater that X gets the smaller Y is , as X cannot be 0 or negative then X must be 1.

[@jamesharmon4994]

While this did turn out to be the answer, I also considered x values between 0 and 1... and quickly abandoned it. 😂

[@JohnSmith-nx7zj]

Why can’t X be negative or 0?

[@paulfillingham2958]

@@JohnSmith-nx7zj if X is 0 you would have Y equal to 0/1 and if X was negative Y would be a negative number divided by a positive number as a squared number is always positive.

[@JohnSmith-nx7zj]

@@paulfillingham2958 ah right yeah, you can establish it pretty easily I see. I thought you meant it was stated in the problem which I didn’t see in the vid.

[@ronaldlau9363]

Or let u=x+1 so that y=(u-1)/u^2=1/u-1/u^2=1/u*(1-1/u), if still not obvious by this point let w=1/u-1/2 making y=(w+1/2)*(1/2-w)=1/4-w^2. [For insight, it might help to first let v=1/u and see y=v*(1-v) before letting w=v-1/2.] {In original post I mistaken wrote u=x-1 instead of u=x+1 or x=u-1, now corrected}

[@maxhagenauer24]

What does that have anything at all to do with the maximum?

[@ronaldlau9363]

@@maxhagenauer24 By having y=1/4-w^2 by a change of variable (or several changes of variable to easily see the motivation), the fact that w^2>=0 means the maximum value of y is 1/4, when w=0, which is when u=2 and x=1.

[@maxhagenauer24]

@ronaldlau9363 So the largest y can be is when w^2 is 0 which makes y = 1/4?

[@leonardobarrera2816]

0:34 I love fractions It makes me understand the world (Relly, not kidding)

[@maxhagenauer24]

y = x/(x+1)^2 u = x+1 y = (u-1)/u^2 y = u^-2(u-1) y = u^(-1)-u^-2 y = 1/u - 1/u^2 Let w = 1/u y = -w^2+w Min = Vertex = -b/2a = -w/2(-w) = 1/2 1/2 = 1/u u = 2 2 = x+1 x = 1 y = 1/(1+1)^2 = 1/4

[@miraj2264]

I'll write [1/(x+1)]^2 as (x+1)^-2 just so it's easier to read in text. y = x*(x+1)^-2 = [(x+1)-1]*(x+1)^-2 = (x+1)*(x+1)^-2 - (1)*(x+1)^-2 = (x+1)^-1 - (x+1)^-2. Let z = (x+1)^-1 ==> y = z - z^2. This is a downward facing quadratic with roots at z = 0 and z = 1. A downward facing quadratic's max occurs at the average of its two roots i.e. z = .5 Plugging back .5 into z ==> .5 = (x+1)^-1 ==> x = 1. Plugging back 1 into x ==> y = 1*(1+1)^-2 = 1/4.

[@ahmedashraf1343]

I have a challenge for you: Compute the sum of k ^ n / 10 ^ k from k = 1 to infinity, where n = 20, Can you do it without differentiating 20 times? Please don't read hints before trying to think about it yourself, and don't go to next hints before thinking about each hint thoroughly, the hints are meant to give you the answer by the third hint Hint I: Compute the sum when n = 3 and 4 Hint II: Can you find a pattern? Hint III: Search for Eulerian numbers

[@leonardobarrera2816]

if the one that got the quafratic formula, he maybe can not belive how much happy it would be seeing this video

[@m.h.6470]

Solution: rewrite (x + 1)² to x² + 2x + 1 factor out an x to get x * (x + 2 + 1/x) You can now cancel the x with the Numerator to get y = 1/(2 + x + 1/x) To get max of y, you now need min of x + 1/x or (x² + 1)/x, that is > -2 If you examine x < -1, you find, that it is < -2 and therefore not valid. If you examine -1 < x < 0, you find, that it too is < -2 and therefore not valid. If you examine 0 < x < 1, you find, that it is > -2 and therefore valid, but quite large and rapidly falling. If you examine 1 < x < 2, you find, that it too is > -2 and therefore valid, but rapidly rising. Conclusion: x = 1 is the minimum for x + 1/x at 2. Therefore the maximum for y = x/(x + 1)² is y = 1/(1 + 1)² = 1/2² = 1/4 at x = 1.

[@RobertssU]

that was nice

[@krabkrabkrab]

I noticed y(x)=y(1/x). Thus, x=1 must be an extremum.

[@PranitSuman]

-0.99999...8?

[@sxkjknjw2]

Would the long way still work? I mean, deriving y and then finding x when the sign changes and then deduce y?

[@SNOWgivemetheid]

yes.

[@Nothingx303]

Crispy but spicy problem 😊

[@aliawahab723]

Prove it in calculus

[@whosFreezy]

why is 2xy, -x = (2y-1)x. Can anyone tell me? around 1:46 in the video

[@mee-tt8mp]

Factor 2xy-x = x(2y-1)

[@rolflangius1119]

(2y-1)x=2y*x-1*x=2xy-x

[@KiyopakaOfficial]

took out of brackets the common multiplier X

[@Eggrics]

There's no comments the comment section. Let me fix this!

[@carultch]

Thank you for coming up with a unique way to fix that, instead of the cliché of the ordinal numbers game.

[@RyanLewis-Johnson-wq6xs]

Find max of y=x/(x+1)^2 max y=0.25=1/4 final answer

[@bjornfeuerbacher5514]

And how did you arrive at that answer?

[@RyanLewis-Johnson-wq6xs]

I did it in my head.

[@bjornfeuerbacher5514]

@@RyanLewis-Johnson-wq6xs But _how_ did you do it in your head?

[@bobross7473]

@@bjornfeuerbacher5514he’s lying

[@CatholicSatan]

Eh? What if x -> -1? Then y is much larger than 1/4.

[@ernstmadsen5526]

My thought too, untill I realized, it goes towards negative infinity at x=-1, so it doesn't affect the max value.

[@RylanceStreet]

If x is negative y is negative so less than 1/4.

[@Viki13]

Numerator would be negative

[@joshvishnu101]

Yeah. Same doubt! What around -1, y goes to infinity

[@discreaminant]

@@joshvishnu101-infinity to be exact so, yeah u got it 😊