How to use differentiation to prove that f is a one to one function Watch

Mesosleepy
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#1
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#1
^^^^^^
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limetang
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I don't know what a formal proof would look like.

But broadly speaking, a differentiable function MUST be continuous, and I think that a continuous function must be one to one.
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TeeEm
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(Original post by Mesosleepy)
^^^^^^
if f'(x) is positive (or negative) within its entire domain then the function is increasing (or decreasing) and therefore 1 to 1
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atsruser
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(Original post by limetang)
I don't know what a formal proof would look like.

But broadly speaking, a differentiable function MUST be continuous, and I think that a continuous function must be one to one.
y=x^2 with the domain [-1,1] is continuous but it's not 1-1 since 1^2=1=(-1)^2
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limetang
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(Original post by atsruser)
y=x^2 with the domain [-1,1] is continuous but it's not 1-1 since 1^2=1=(-1)^2
Excellent point.
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limetang
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Oh, got it. A continuous function with no turning points is 1 to 1 (I think). so f''(x) = 0 should have no so solutions.
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donutellme
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Differentiate and see if the gradient flips signs. A 1-1 function can only be increasing or decreasing.
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atsruser
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(Original post by limetang)
Excellent point.
You were right about the differentiable => continuous bit though. (The implication doesn't go the other way though - for example, |x| is continuous at 0 but not differentiable there)
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Lord of the Flies
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Gosh a lot of false assumptions here.

In general you cannot use differentiation to prove injectivity, because injectivity has absolutely nothing to do with continuity let alone differentiability. Even assuming the function is continuous and injective, it need not be differentiable (heck you can even get it not to be differentiable at uncountably many points).
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Implication
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(Original post by Lord of the Flies)
Gosh a lot of false assumptions here.

In general you cannot use differentiation to prove injectivity, because injectivity has absolutely nothing to do with continuity let alone differentiability. Even assuming the function is continuous and injective, it need not be differentiable (heck you can even get it not to be differentiable at uncountably many points).
Well a strictly increasing continuous function will always be injective, won't it? And if the derivative exists and is positive everywhere, then the function is continuous and strictly increasing and hence injective?

Am I wrong?
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Lord of the Flies
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(Original post by Implication)
Well a strictly increasing continuous function will always be injective, won't it? And if the derivative exists and is positive everywhere, then the function is continuous and strictly increasing and hence injective?
You're not wrong - I was merely saying that differentiability is a very heavy assumption indeed, and that even if your function is continuous and strictly increasing, differentiation will still not work (in general). I understood the OP as "I want to show a function is injective - how can I use differentiation to achieve this?"
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