dy/dx = 0

[anantchowdhary]

If dy/dx=0,
for all x in the function y= f(x)'s domain,then
how can we say... only by using limits, that the function is a constant one?

[morphism]

Only by using limits? What's wrong with the way this is usually proved, i.e. via the mean value theorem?

[anantchowdhary]

the mean value theorem here....im not sure...cuz we might get a limit of the function staying constant..but i dont see how the function is EXACTLY constant, being proved by only using the first principle...

thanks

[GoldPheonix]

If dy/dx=0,
for all x in the function y= f(x)'s domain,then
how can we say... only by using limits, that the function is a constant one?
As far as a limit definition:

dy/dx = lim[ F(x+h) - F(x)]/h

In order for that to be true, then:

F(x+h) - F(x) = 0

=>

F(x+h) = F(x)

Which means that you have an equation which does not vary with respect to x, otherwise there would be something left over from that subtraction. The only equation which does not vary with respect to x is a function that is constant, relative to x.

[anantchowdhary]

cant F(x+h)-F(x) be <<h ...I dont think it can as h is a dynamic variable,which is infinitesimally small...so am i correct
Please correct me if im wrong
thanks

[nicksauce]

If (A-B)/C = 0, and C is not 0, then it has to follow that A=B.

[anantchowdhary]

But this is a limit

i mean lim(a--->b)a-b/c=0

it duznt mean a=b!

[WarPhalange]

But this is a limit

i mean lim(a--->b)a-b/c=0

it duznt mean a=b!

No, it's lim(c -> 0) of (a-b)/c

[anantchowdhary]

ok...i got my mistake...but cant a-b be something like c^2

then the limit still is 0 but we cant say a=b...can we?

[HallsofIvy]

In order that this be true, f must be differentiable at every point in the interval. Your limit calculations are not using that.

[morphism]

dy/dx = lim[ F(x+h) - F(x)]/h

In order for that to be true, then:

F(x+h) - F(x) = 0

This is not true. Take F(x)=x^2 for instance, and apply the limit definition at x=0:

F'(0) = \lim_{h \to 0} \frac{F(0+h) - F(0)}{h} = \lim_{h \to 0} \frac{h^2}{h} = 0.

But F(0+h)-F(0)=h^2 is not zero for all h.

This is why you need to use a deeper result like the mean value theorem. It tells you that if x is not equal to y, then there is a c such that

\frac{F(x) - F(y)}{x - y} = F'(c) = 0 \implies F(x) = F(y).

[Note: I'm assuming that F is differentiable everywhere so that the hypotheses of the MVT are satisfied.]

[anantchowdhary]

Is it correct to simply say ..that in a function we cant have an infinitesimal so the limit of the derivative is not zero unitl it IS constant
or like we have a function f(x)=a(x)-b(x)
if a`(x)=b`(x) for all x

then a(x)=b(x) necessarily using the above argument(infinitesimals)

[morphism]

Is it correct to simply say ..that in a function we cant have an infinitesimal so the limit of the derivative is not zero unitl it IS constant
That doesn't really make any sense.

[anantchowdhary]

ok...

[anantchowdhary]

So how would we exactly prove the required result using only limits and logic?

thanks

[HallsofIvy]

Use limits (and logic) to prove the mean value theorem! One point that has been made repeatedly here is that you can't use limits at a single point because saying that y= 0 identically (on some interval) is a property of that interval, not a single point.

[anantchowdhary]

ok...i get your point:smile:

but what if we have a function like...

f(x+y)=f(x)+y^3
and it is provided that the function is continuous and differentiable
so this gives us

f(x+h)=f(x)+h^3

or

\frac {f(x+h)-f(x)}{h}=h^2


\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}=0

but i dont think the function is constant....
is there anything wrong in the function?

[HallsofIvy]

ok...i get your point:smile:

but what if we have a function like...

f(x+y)=f(x)+y^3
What reason do you have to believe such a function exists? If so, taking x= 0, f(y)= y3. But then
f(x+y)= (x+y)^3= x^3+ 3x^2y+ 3xy^2+ y^3= f(x)+ 3x^2y+ 3xy^2+ y^3\ne f(x)+ y^3
a contradiction.

and it is provided that the function is continuous and differentiable
so this gives us

f(x+h)=f(x)+h^3

or

\frac {f(x+h)-f(x)}{h}=h^2


\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}=0

but i dont think the function is constant....
is there anything wrong in the function?

[anantchowdhary]

I agree.. that the function wont exist...

but,using the MVT arent we still just proving the tendency not the EXACT equality?
i know this is irritating

but

thanks anyways

[morphism]

but,using the MVT arent we still just proving the tendency not the EXACT equality?

No. The MVT is an exact equality. See: http://planetmath.org/encyclopedia/MeanValueTheorem.html.

[anantchowdhary]

:rolleyes: Well,I didnt mean to say that MVT isnt an exact equality

[HallsofIvy]

then perhaps you should explain what you mean by "tendency" and "exact equality".

[anantchowdhary]

I just got a bit confused about the tendency of the function being a constant..i mean to say that i thought the function woould vary extremely slowly...my mistake!