Let's derivate 3
True or false :
Let f ( x ) : ( 0 , ∞ ) ⟶ R such that f ( x ) = ln ( 1 + x ) ∀ x ∈ ( 0 , ∞ ) , then there exists x ∈ ( 0 , ∞ ) such that f ( x ) ≥ x .
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1 solution
haha, what a original shape for solving problems,Comrade,haha... thank you for your solution,interesting...
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I like to see things whenever I can, Comrade, rather than relying blindly on algebra or analysis.
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I started with Peano axioms and there I finished... haha, I was too hard...
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@Guillermo Templado – It's good to be able to do it the hard way too, but, in my humble opinion, we should always start with an attempt to see things and grasp them intuitively. I learned most of my math from Soviet applied math books (mostly from the Steklov Institute in Leningrad, ЛОМИ), and those Comrades took a wonderfully practical approach.
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@Otto Bretscher – I have learned a little from here and little bit from there, I have had a lot of leaders: books, and teachers, I like Maths due to its logic... this is true due to this and for here we can also we get that, this is not true due to... and If the logic is good there a lot of ways to arrive the right mark with art
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@Guillermo Templado – Here is an analytic solution, using the Mean Value Theorem, to satisfy the formalists.
For any positive x we have x f ( x ) = x − 0 f ( x ) − f ( 0 ) = f ′ ( c ) = c + 1 1 < 1 for some c between 0 and x . Thus f ( x ) < x for all positive x .
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@Otto Bretscher – ok, I'm going to post my another solution. It's very similar...
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@Guillermo Templado – Let g ( x ) = ln ( 1 + x ) − x ∀ x ∈ [ 0 , ∞ ) ⇒ g ( 0 ) = 0 and g ′ ( x ) = x + 1 1 − 1 < 0 ∀ x ∈ ( 0 , ∞ ) this says us that g is strictly decreasing ∀ x ∈ ( 0 , ∞ ) ⇒ g ( x ) = f ( x ) − x < 0 ∀ x ∈ ( 0 , ∞ ) ... and x g ( x ) − g ( 0 ) = g ′ ( c ) with x , c ∈ ( 0 , ∞ ) ⇒ g ( x ) − g ( 0 ) < 0 . . .
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@Guillermo Templado – Yes, exactly, that is really the same solution. As you know, the fact that a function with a negative derivative is strictly decreasing follows from the Mean Value Theorem.
Just a small correction to show that I read your solution carefully: g ′ ( x ) < 0 for x > 0 but not for x = 0 .
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@Otto Bretscher – For example.... you can also use the definition of limit too. If f '(c) < 0 then there exists a neighborhood of c where f ( x ) = f ( c ) + f ′ ( c ) ( x − c ) + o ( ∣ x − c ∣ ) ⇒ f is strictly decreasing. at that neighborhood
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@Guillermo Templado – I'm going to play the formalist now: I don't quite see how that neighborhood argument proves that the function is decreasing in the neighborhood.
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@Otto Bretscher – Pay attention, I think I'm right x → c lim x − c f ( x ) − f ( c ) = f ′ ( c ) ⇒ x → c lim x − c f ( x ) − f ( c ) − f ′ ( c ) = 0 ⇒ f ( x ) − f ( c ) − f ′ ( c ) ( x − c ) = o ( ∣ x − c ∣ ) Then,we can ignore o(|x - c|) in that neighborhood
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@Guillermo Templado – I do pay attention, but I don't see it. You merely prove that f ( x ) < f ( c ) when x > c and vice versa, but that does not imply that the function is decreasing in the neighborhood. The claim "If f ′ ( c ) < 0 , then f ( x ) is decreasing in some neighborhood of c " is not correct, I believe.
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@Otto Bretscher – ok, If f '(c) < 0 , then if x > c , f(x) = f(c) + f '(c)(x -c) + o(|x -c|) < f(c) in that neighborhood and
if x < c then f(x) = f(c) + f '(c) (x - c) + o(|x - c|) > f(c) in that neighborhood
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@Guillermo Templado – Yes, indeed, but that does not imply that the function is decreasing in that neighborhood ;) You may have points a < b < c in the neighborhood such that f ( a ) < f ( b ) .
I can give you an example of a function with f ′ ( c ) < 0 that is not decreasing in any neighborhood of c , if you insist.
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@Otto Bretscher – Si f has a continuous derivate in this neighborhood, this works and it say us it is strictly decreasing. Another thing is if f doesn't have a continuous derivative. You are saying me this, I think
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@Guillermo Templado – Sure, if f ′ ( x ) is assumed to be continuous and f ′ ( c ) < 0 , then f ′ ( x ) will be negative on some neighborhood of c and therefore f ( x ) will be strictly decreasing on that neighborhood. To prove this, you need the Mean Value Theorem (or equivalent); you cannot get it from the limit definition of the derivative alone.
@Otto Bretscher – Yes,Give me one example, please
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@Guillermo Templado – Take f ( x ) = 2 x 2 sin ( x 1 ) − x if x is nonzero and f ( 0 ) = 0 . We find that f ′ ( 0 ) = − 1 but f ( x ) isn't decreasing in any neighborhood of 0 since f ′ ( ( 2 k + 1 ) π 1 ) = 1 for all integers k .
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@Otto Bretscher – hum, thank you, very interesting, I'll think this function later,,,.
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@Guillermo Templado – I always discuss this function briefly in my introductory calculus classes as an example of a differentiable function whose derivative fails to be continuous.
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@Otto Bretscher – Hum, very interesting too, because if f : R n ⟶ R has continuous its first partials derivatives then f is a differentiable function, and you are proving me than there exists a function differentiable whose derivative fails to be continuous.Hum,very interesting, I'll continue thinking about this, you are "tremendo" tremendous, thundering
@Otto Bretscher – I wrote it good, didn't I?
@Guillermo Templado – I'm approaching retirement (in Cuba?), and I see many of my friends and neighbors getting a bit bored and inert after retirement. I think that will not happen to me, in part thanks to brilliant.org ;)
y = x is the tangent to y = ln ( 1 + x ) at the origin. Since the graph of ln ( 1 + x ) is concave, the graph is below the tangent, so that ln ( 1 + x ) < x for all positive x (and also for all − 1 < x < 0 ).