Check my proof please
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Let $f,g:mathbb{R} -> mathbb{R}$ be two continuous,differentiable functions.Prove that if $f'(x)=g'(x)$ then $f(x)-g(x)=C$,where C is a real constant.
My proof: Suppose $f(x)-g(x) neq C$.
By differentiating the above we get that $f'(x)-g'(x) neq 0$,false.
I am not really sure if by differentiating $ neq$ still holds and this is why I am asking for your help.
functions derivatives
asked Nov 18 at 10:41
JustAnAmateur
92
add a comment |
up vote
-2
down vote
favorite
Let $f,g:mathbb{R} -> mathbb{R}$ be two continuous,differentiable functions.Prove that if $f'(x)=g'(x)$ then $f(x)-g(x)=C$,where C is a real constant.
My proof: Suppose $f(x)-g(x) neq C$.
By differentiating the above we get that $f'(x)-g'(x) neq 0$,false.
I am not really sure if by differentiating $ neq$ still holds and this is why I am asking for your help.
functions derivatives
asked Nov 18 at 10:41
JustAnAmateur
92
It does hold, in this situation. Unfortunately, that's exactly the result that you're trying to prove. For a hint, notice that the result is false if the domain is not connected, so your proof will have to use the connectedness of $mathbb{R}$ at some point.
– user3482749
Nov 18 at 10:44
I see. So how should I approach proving this then?
– JustAnAmateur
Nov 18 at 10:46
Since differentiation is linear, (f-g)' = f' - g' = 0. Thus, you just need to prove that if a function has zero derivative, then it is constant.
– user3482749
Nov 18 at 10:55
The result is true if $;f'(x)=g'(x);$ for some $;xin I;,;;I = $ a *non-trivial (open) interval*. Otherwise it is false, of course. For the proof: do you already know that $;h'(x)=0;$ for all values in some non-trivial interval , iff $;h;$ is a constant on that interval ?
– DonAntonio
Nov 18 at 10:55
@JustAnAmateur Your statement is a simple consequence of the mean value theorem. E.g, see the following Wikipedia link - en.wikipedia.org/wiki/Mean_value_theorem#A_simple_application
– Marian G.
Nov 18 at 13:57
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let $f,g:mathbb{R} -> mathbb{R}$ be two continuous,differentiable functions.Prove that if $f'(x)=g'(x)$ then $f(x)-g(x)=C$,where C is a real constant.
My proof: Suppose $f(x)-g(x) neq C$.
By differentiating the above we get that $f'(x)-g'(x) neq 0$,false.
I am not really sure if by differentiating $ neq$ still holds and this is why I am asking for your help.
functions derivatives
asked Nov 18 at 10:41
JustAnAmateur
92
Let $f,g:mathbb{R} -> mathbb{R}$ be two continuous,differentiable functions.Prove that if $f'(x)=g'(x)$ then $f(x)-g(x)=C$,where C is a real constant.
My proof: Suppose $f(x)-g(x) neq C$.
By differentiating the above we get that $f'(x)-g'(x) neq 0$,false.
I am not really sure if by differentiating $ neq$ still holds and this is why I am asking for your help.
functions derivatives
functions derivatives
asked Nov 18 at 10:41
JustAnAmateur
92
asked Nov 18 at 10:41
JustAnAmateur
92
asked Nov 18 at 10:41
JustAnAmateur
92
asked Nov 18 at 10:41
JustAnAmateur
92
asked Nov 18 at 10:41
JustAnAmateur
92
92
It does hold, in this situation. Unfortunately, that's exactly the result that you're trying to prove. For a hint, notice that the result is false if the domain is not connected, so your proof will have to use the connectedness of $mathbb{R}$ at some point.
– user3482749
Nov 18 at 10:44
I see. So how should I approach proving this then?
– JustAnAmateur
Nov 18 at 10:46
Since differentiation is linear, (f-g)' = f' - g' = 0. Thus, you just need to prove that if a function has zero derivative, then it is constant.
– user3482749
Nov 18 at 10:55
The result is true if $;f'(x)=g'(x);$ for some $;xin I;,;;I = $ a *non-trivial (open) interval*. Otherwise it is false, of course. For the proof: do you already know that $;h'(x)=0;$ for all values in some non-trivial interval , iff $;h;$ is a constant on that interval ?
– DonAntonio
Nov 18 at 10:55
@JustAnAmateur Your statement is a simple consequence of the mean value theorem. E.g, see the following Wikipedia link - en.wikipedia.org/wiki/Mean_value_theorem#A_simple_application
– Marian G.
Nov 18 at 13:57
add a comment |
It does hold, in this situation. Unfortunately, that's exactly the result that you're trying to prove. For a hint, notice that the result is false if the domain is not connected, so your proof will have to use the connectedness of $mathbb{R}$ at some point.
– user3482749
Nov 18 at 10:44
I see. So how should I approach proving this then?
– JustAnAmateur
Nov 18 at 10:46
Since differentiation is linear, (f-g)' = f' - g' = 0. Thus, you just need to prove that if a function has zero derivative, then it is constant.
– user3482749
Nov 18 at 10:55
The result is true if $;f'(x)=g'(x);$ for some $;xin I;,;;I = $ a *non-trivial (open) interval*. Otherwise it is false, of course. For the proof: do you already know that $;h'(x)=0;$ for all values in some non-trivial interval , iff $;h;$ is a constant on that interval ?
– DonAntonio
Nov 18 at 10:55
@JustAnAmateur Your statement is a simple consequence of the mean value theorem. E.g, see the following Wikipedia link - en.wikipedia.org/wiki/Mean_value_theorem#A_simple_application
– Marian G.
Nov 18 at 13:57
It does hold, in this situation. Unfortunately, that's exactly the result that you're trying to prove. For a hint, notice that the result is false if the domain is not connected, so your proof will have to use the connectedness of $mathbb{R}$ at some point.
– user3482749
Nov 18 at 10:44
It does hold, in this situation. Unfortunately, that's exactly the result that you're trying to prove. For a hint, notice that the result is false if the domain is not connected, so your proof will have to use the connectedness of $mathbb{R}$ at some point.
– user3482749
Nov 18 at 10:44
I see. So how should I approach proving this then?
– JustAnAmateur
Nov 18 at 10:46
I see. So how should I approach proving this then?
– JustAnAmateur
Nov 18 at 10:46
Since differentiation is linear, (f-g)' = f' - g' = 0. Thus, you just need to prove that if a function has zero derivative, then it is constant.
– user3482749
Nov 18 at 10:55
Since differentiation is linear, (f-g)' = f' - g' = 0. Thus, you just need to prove that if a function has zero derivative, then it is constant.
– user3482749
Nov 18 at 10:55
The result is true if $;f'(x)=g'(x);$ for some $;xin I;,;;I = $ a *non-trivial (open) interval*. Otherwise it is false, of course. For the proof: do you already know that $;h'(x)=0;$ for all values in some non-trivial interval , iff $;h;$ is a constant on that interval ?
– DonAntonio
Nov 18 at 10:55
The result is true if $;f'(x)=g'(x);$ for some $;xin I;,;;I = $ a *non-trivial (open) interval*. Otherwise it is false, of course. For the proof: do you already know that $;h'(x)=0;$ for all values in some non-trivial interval , iff $;h;$ is a constant on that interval ?
– DonAntonio
Nov 18 at 10:55
@JustAnAmateur Your statement is a simple consequence of the mean value theorem. E.g, see the following Wikipedia link - en.wikipedia.org/wiki/Mean_value_theorem#A_simple_application
– Marian G.
Nov 18 at 13:57
@JustAnAmateur Your statement is a simple consequence of the mean value theorem. E.g, see the following Wikipedia link - en.wikipedia.org/wiki/Mean_value_theorem#A_simple_application
– Marian G.
Nov 18 at 13:57
add a comment |
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It does hold, in this situation. Unfortunately, that's exactly the result that you're trying to prove. For a hint, notice that the result is false if the domain is not connected, so your proof will have to use the connectedness of $mathbb{R}$ at some point.
– user3482749
Nov 18 at 10:44
I see. So how should I approach proving this then?
– JustAnAmateur
Nov 18 at 10:46
Since differentiation is linear, (f-g)' = f' - g' = 0. Thus, you just need to prove that if a function has zero derivative, then it is constant.
– user3482749
Nov 18 at 10:55
The result is true if $;f'(x)=g'(x);$ for some $;xin I;,;;I = $ a *non-trivial (open) interval*. Otherwise it is false, of course. For the proof: do you already know that $;h'(x)=0;$ for all values in some non-trivial interval , iff $;h;$ is a constant on that interval ?
– DonAntonio
Nov 18 at 10:55
@JustAnAmateur Your statement is a simple consequence of the mean value theorem. E.g, see the following Wikipedia link - en.wikipedia.org/wiki/Mean_value_theorem#A_simple_application
– Marian G.
Nov 18 at 13:57