Actually, we have for every x ( x^2 - 2x - 3 )/( 2x^2 + 2x + 1 ) + 4 = ( 3x + 1 )^2/( 2x^2 + 2x + 1 ). This implies that the function reaches the minimum - 4 when x = - 1/3.
I wondered how to come up with this approach intuitively (at least from my perspective, anyone has a different intuition of course) and this is what I came up with: Set x²-2x-3 / 2x²+ 2x +1 >= A Now find an A that there is exactly one solution in x. I do now see, that the video did something very, very similar, if not the same. But the little twist, that helped me was to use "A" instead of "y", thus my mindset was set on a constant, we need to find and not on a function we need to treat in any way.
Ok, but in this case if you just derivate ... N'D-ND' = 0 is a very simple quadratic equation that gives you much more info about the function ... f(-infinity)=f(+infinity)=1/2, N factoris. as (x+1)(x-3) function has zeroes at -1 and 3, D unfactoris. function has no pole f'=0 has 2 solutions at -2 and -1/3 f(-2)=1 is the maximum and f(-1/3)=-4 is the minimum. And all these informations requires no more calculus than what is in the video ...
I think it is no longer needed to find an x for which the value -4 is attained since the method used, along with noting that the denominator is always non-zero, guarantees such an x exists (we had a quadratic in x to start with and we precisely imposed that its discriminant must be non-zero in order for it to have a solution x).
@@leif1075 no, I do not, unfortunately. the denominator cannot be 0 for real x. this fact was also proved in the video (for example, the discriminant is strictly negative, so no real roots exist, but the video shows a proof based on forming a square)
@@leif1075 It does imply we are discussing real x's only since we are asked to compute the 'minimum' of an expression in x. If we allow x to take complex non-real values, then plugging, say, i, gives a value that is non-real, and complex numbers are not ordered. Also, by inspecting the proof in the video, it can be noticed that real numbers are assumed (it is argued that the denominator is always > 0 by using the fact that the square of a real number is always >=0, something which is not necessarily true for complex numbers, where the square is not even always real). EDIT: While the problem statement is indeed somehow incomplete since it does not explicitly say we are discussing real x's, this is usually the case for problems like this, where we are asked to compute the minimum of an expression in x, since a 'minimum' implies that an ordering exists on the set where we allow our expression to take values from. It is true, however, that one could try solving the following: find the minimum of the given expression on the set of complex x's for which the expression (exists and) takes real values. Is this what you were saying?
@@bait6652 what do you mean?? .5 is just a constant there's no variable term at all..did you mean to write x squared times 0.5 you mean? Otherwise that makes no sense..sorry don't knkw what you mean..and why woukd you think of that..why not rewrite the numerator as (x-3)(× +1) like I did?
Sure, but sometimes brute force calculus is not necessary longer ... it took me much less than 6 minutes to derivate the function and find both the min and the max ^^
Actually what is happening here is nothing but finding the domain of the inverse function. And that can be done due to the nature of the function being quadratic.
Actually, we have for every x
( x^2 - 2x - 3 )/( 2x^2 + 2x + 1 ) + 4 = ( 3x + 1 )^2/( 2x^2 + 2x + 1 ).
This implies that the function reaches the minimum - 4 when x = - 1/3.
Nice
Excuse me where did that 4 come from?? Whybwoukd you add4..that random and out of nowhere...
@@leif1075
The similar way in the video. This is just verification that - 4 is minimum.
Can you plz solve this question a^1/3 +b^1/3 +c^1/3 = (2^1/3 - 1)^1/3 . Find a+b+c
I wondered how to come up with this approach intuitively (at least from my perspective, anyone has a different intuition of course) and this is what I came up with:
Set x²-2x-3 / 2x²+ 2x +1 >= A
Now find an A that there is exactly one solution in x. I do now see, that the video did something very, very similar, if not the same.
But the little twist, that helped me was to use "A" instead of "y", thus my mindset was set on a constant, we need to find and not on a function we need to treat in any way.
Ok, but in this case if you just derivate ... N'D-ND' = 0 is a very simple quadratic equation that gives you much more info about the function ...
f(-infinity)=f(+infinity)=1/2, N factoris. as (x+1)(x-3) function has zeroes at -1 and 3, D unfactoris. function has no pole
f'=0 has 2 solutions at -2 and -1/3 f(-2)=1 is the maximum and f(-1/3)=-4 is the minimum.
And all these informations requires no more calculus than what is in the video ...
I think it is no longer needed to find an x for which the value -4 is attained since the method used, along with noting that the denominator is always non-zero, guarantees such an x exists (we had a quadratic in x to start with and we precisely imposed that its discriminant must be non-zero in order for it to have a solution x).
Not necessarily true though if 2x squared plus 2x equals negstive 1 then the denominator would be zero..see what I mean?
@@leif1075 no, I do not, unfortunately. the denominator cannot be 0 for real x. this fact was also proved in the video (for example, the discriminant is strictly negative, so no real roots exist, but the video shows a proof based on forming a square)
@@BucifalulR OK but why linit yourself to real roots..the video never says only look for real solutions..now I hope you see..
@@leif1075 It does imply we are discussing real x's only since we are asked to compute the 'minimum' of an expression in x. If we allow x to take complex non-real values, then plugging, say, i, gives a value that is non-real, and complex numbers are not ordered. Also, by inspecting the proof in the video, it can be noticed that real numbers are assumed (it is argued that the denominator is always > 0 by using the fact that the square of a real number is always >=0, something which is not necessarily true for complex numbers, where the square is not even always real). EDIT: While the problem statement is indeed somehow incomplete since it does not explicitly say we are discussing real x's, this is usually the case for problems like this, where we are asked to compute the minimum of an expression in x, since a 'minimum' implies that an ordering exists on the set where we allow our expression to take values from. It is true, however, that one could try solving the following: find the minimum of the given expression on the set of complex x's for which the expression (exists and) takes real values. Is this what you were saying?
Can you plz solve this question a^1/3 +b^1/3 +c^1/3 = (2^1/3 - 1)^1/3 . Find a+b+c
If you rewrite the equation as 0.5-(3x+3.5)/(2x²+2x +1) if makes life a lot simpler, both with and without calculus.
how? does it avoid discr or deriv?
You can't though. What happened tonthe x squared and 2 x terms in the nhmerstor..you can't just erase them..
@@leif1075 x^2 is part of 0.5
@@bait6652 what do you mean?? .5 is just a constant there's no variable term at all..did you mean to write x squared times 0.5 you mean? Otherwise that makes no sense..sorry don't knkw what you mean..and why woukd you think of that..why not rewrite the numerator as (x-3)(× +1) like I did?
@@leif1075 his 0.5 is not part of the numerator
Can you plz solve this question a^1/3 +b^1/3 +c^1/3 = (2^1/3 - 1)^1/3 . Find a+b+c
thank you
wow thank you
We may test that
f ' (x) = 0 when x = - 1/3
zebi
Sure, but sometimes brute force calculus is not necessary longer ... it took me much less than 6 minutes to derivate the function and find both the min and the max ^^
I know this is the quadratic method. Anyway, thx for sharing.
Tks.
Very nice!
Actually what is happening here is nothing but finding the domain of the inverse function. And that can be done due to the nature of the function being quadratic.
cool
There MUST be another way to solve..I don't see why anyone would ever think of this at ALL..it's all out of nowhere
No, every time you see min/max questions involving 2nd powered terms, use this method. It's much more fun this way
Derivative: what?
Derivative in Calculus be like: Nah, I didn't see the title.