Simple Derivative Proof? [on hold]Proof of derivative of invertible functionHow to make a piecewise function differentiable?The Differentiability of a Piecewise FunctionProof of Lipschitz continuousA question about existence of derivative of function at ZeroQuestion about derivative proof :)Derivative of a power functionProperties of the mean value theorem derivativeIs a function strictly increasing if its derivative is positive at all point but critical points?Limit definition of a derivative proof?
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Simple Derivative Proof? [on hold]
Proof of derivative of invertible functionHow to make a piecewise function differentiable?The Differentiability of a Piecewise FunctionProof of Lipschitz continuousA question about existence of derivative of function at ZeroQuestion about derivative proof :)Derivative of a power functionProperties of the mean value theorem derivativeIs a function strictly increasing if its derivative is positive at all point but critical points?Limit definition of a derivative proof?
$begingroup$
Suppose that $f$ is a function with the following properties:
$f$ is differentiable everywhere
$f(x+y) = f(x)f(y), f(0)≠0$
$f'(0)=1$
$f(0)=1$
How do you prove that $f(x)>0$ for all values of $x$? Please explain the steps, as I am in an intro class.
algebra-precalculus functions derivatives
edited May 5 at 1:37
YuiTo Cheng
3,21161145
asked May 4 at 23:45
MaceyMacey
132
New contributor
Macey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by YuiTo Cheng, RRL, GNUSupporter 8964民主女神 地下教會, TheSimpliFire, user21820 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YuiTo Cheng, RRL, GNUSupporter 8964民主女神 地下教會, TheSimpliFire, user21820
add a comment |
$begingroup$
Suppose that $f$ is a function with the following properties:
$f$ is differentiable everywhere
$f(x+y) = f(x)f(y), f(0)≠0$
$f'(0)=1$
$f(0)=1$
How do you prove that $f(x)>0$ for all values of $x$? Please explain the steps, as I am in an intro class.
algebra-precalculus functions derivatives
edited May 5 at 1:37
YuiTo Cheng
3,21161145
asked May 4 at 23:45
MaceyMacey
132
New contributor
Macey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by YuiTo Cheng, RRL, GNUSupporter 8964民主女神 地下教會, TheSimpliFire, user21820 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YuiTo Cheng, RRL, GNUSupporter 8964民主女神 地下教會, TheSimpliFire, user21820
$begingroup$
Here is a helpful clue. For any base "b". $b^x+y = b^xb^y$
$endgroup$
– Q the Platypus
May 4 at 23:48
$begingroup$
Incidentally, the assumption that $f$ is differentiable everywhere is redundant. Just from the assumption that $f$ is differentiable at $x=0$ we then get $lim_hto 0 fracf(x+h)-f(x)h = lim_hto 0 fracf(x)f(h)-f(x)h = f(x) lim_hto 0 fracf(h)-1h = f(x) f'(0) = f(x)$ exists for every $x$.
$endgroup$
– Daniel Schepler
May 5 at 5:24
add a comment |
$begingroup$
Suppose that $f$ is a function with the following properties:
$f$ is differentiable everywhere
$f(x+y) = f(x)f(y), f(0)≠0$
$f'(0)=1$
$f(0)=1$
How do you prove that $f(x)>0$ for all values of $x$? Please explain the steps, as I am in an intro class.
algebra-precalculus functions derivatives
edited May 5 at 1:37
YuiTo Cheng
3,21161145
asked May 4 at 23:45
MaceyMacey
132
New contributor
Macey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Suppose that $f$ is a function with the following properties:
$f$ is differentiable everywhere
$f(x+y) = f(x)f(y), f(0)≠0$
$f'(0)=1$
$f(0)=1$
How do you prove that $f(x)>0$ for all values of $x$? Please explain the steps, as I am in an intro class.
algebra-precalculus functions derivatives
algebra-precalculus functions derivatives
edited May 5 at 1:37
YuiTo Cheng
3,21161145
asked May 4 at 23:45
MaceyMacey
132
New contributor
Macey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited May 5 at 1:37
YuiTo Cheng
3,21161145
asked May 4 at 23:45
MaceyMacey
132
New contributor
Macey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited May 5 at 1:37
YuiTo Cheng
3,21161145
edited May 5 at 1:37
YuiTo Cheng
3,21161145
edited May 5 at 1:37
YuiTo Cheng
3,21161145
3,21161145
asked May 4 at 23:45
MaceyMacey
132
New contributor
Macey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 4 at 23:45
MaceyMacey
132
asked May 4 at 23:45
MaceyMacey
132
132
New contributor
Macey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Macey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by YuiTo Cheng, RRL, GNUSupporter 8964民主女神 地下教會, TheSimpliFire, user21820 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YuiTo Cheng, RRL, GNUSupporter 8964民主女神 地下教會, TheSimpliFire, user21820
put on hold as off-topic by YuiTo Cheng, RRL, GNUSupporter 8964民主女神 地下教會, TheSimpliFire, user21820 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YuiTo Cheng, RRL, GNUSupporter 8964民主女神 地下教會, TheSimpliFire, user21820
$begingroup$
Here is a helpful clue. For any base "b". $b^x+y = b^xb^y$
$endgroup$
– Q the Platypus
May 4 at 23:48
$begingroup$
Incidentally, the assumption that $f$ is differentiable everywhere is redundant. Just from the assumption that $f$ is differentiable at $x=0$ we then get $lim_hto 0 fracf(x+h)-f(x)h = lim_hto 0 fracf(x)f(h)-f(x)h = f(x) lim_hto 0 fracf(h)-1h = f(x) f'(0) = f(x)$ exists for every $x$.
$endgroup$
– Daniel Schepler
May 5 at 5:24
add a comment |
$begingroup$
Here is a helpful clue. For any base "b". $b^x+y = b^xb^y$
$endgroup$
– Q the Platypus
May 4 at 23:48
$begingroup$
Incidentally, the assumption that $f$ is differentiable everywhere is redundant. Just from the assumption that $f$ is differentiable at $x=0$ we then get $lim_hto 0 fracf(x+h)-f(x)h = lim_hto 0 fracf(x)f(h)-f(x)h = f(x) lim_hto 0 fracf(h)-1h = f(x) f'(0) = f(x)$ exists for every $x$.
$endgroup$
– Daniel Schepler
May 5 at 5:24
$begingroup$
Here is a helpful clue. For any base "b". $b^x+y = b^xb^y$
$endgroup$
– Q the Platypus
May 4 at 23:48
$begingroup$
Here is a helpful clue. For any base "b". $b^x+y = b^xb^y$
$endgroup$
– Q the Platypus
May 4 at 23:48
$begingroup$
Incidentally, the assumption that $f$ is differentiable everywhere is redundant. Just from the assumption that $f$ is differentiable at $x=0$ we then get $lim_hto 0 fracf(x+h)-f(x)h = lim_hto 0 fracf(x)f(h)-f(x)h = f(x) lim_hto 0 fracf(h)-1h = f(x) f'(0) = f(x)$ exists for every $x$.
$endgroup$
– Daniel Schepler
May 5 at 5:24
$begingroup$
Incidentally, the assumption that $f$ is differentiable everywhere is redundant. Just from the assumption that $f$ is differentiable at $x=0$ we then get $lim_hto 0 fracf(x+h)-f(x)h = lim_hto 0 fracf(x)f(h)-f(x)h = f(x) lim_hto 0 fracf(h)-1h = f(x) f'(0) = f(x)$ exists for every $x$.
$endgroup$
– Daniel Schepler
May 5 at 5:24
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For any $xinBbb R$, we have $f(x)=f(frac x2+frac x2)=left(f(x/2)right)^2geq0$, so we know $f(x)geq0$ for all $xinBbb R$. Now suppose $f(x_0)=0$. Then $f(0)=f(x_0+(-x_0))=f(x_0)f(-x_0)=0$, which is a contradiction. Thus $f(x)>0$ for all $x$.
answered May 4 at 23:52
ClaytonClayton
20.1k33488
$endgroup$
add a comment |
$begingroup$
The first step when thinking about a problem like this would be to ask "What would happen if $f(z) = 0$ for some value?". Well since $f(x + y) = f(x)f(y)$ that would mean you could subtract z from any number to find a $f(x)f(z)$ pair that would have to equal zero. So all the numbers would have to be zero, which is contradicted by the questions requirement that $f(0)=1$.
You can make a similar argument for negative numbers. Any positive number times a negative number is a negative number. So if there was a number that make $f(x)$ negative then you could always find a $f(x)f(y)$ that would also be negative and force f(0) to be negative which we know it is not.
answered May 5 at 0:08
Q the PlatypusQ the Platypus
2,9821136
$endgroup$
$begingroup$
I'm not sure what you mean when you say you can make a similar argument for negative numbers. Do you mean that saying something like f(z) = -1 for some value then you would need an x+y pair that = -1? Sorry, I am not good at understanding proofs.
$endgroup$
– Macey
May 5 at 0:25
add a comment |
$begingroup$
If you diffentiate the equality wrt. $x$, you get $f'(x+y)=f'(x)f(y)$, so for $x=0$ you get $$f'(y)=f(y)$$
You may know that the solutions of this differential equation are of the form $lambda e^x$. Applying $f'(0)=1$ you get $f(x)=e^x$ as a unique solution, and it is indeed $>0$
answered May 4 at 23:58
elidiotelidiot
1,27814
$endgroup$
$begingroup$
How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
$endgroup$
– Macey
May 5 at 0:03
$begingroup$
$f'=f$ is a differential equation, which I solve to answer the question. Read until the end, I can conclude by saying that the exponential is $>0$
$endgroup$
– elidiot
May 5 at 0:05
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For any $xinBbb R$, we have $f(x)=f(frac x2+frac x2)=left(f(x/2)right)^2geq0$, so we know $f(x)geq0$ for all $xinBbb R$. Now suppose $f(x_0)=0$. Then $f(0)=f(x_0+(-x_0))=f(x_0)f(-x_0)=0$, which is a contradiction. Thus $f(x)>0$ for all $x$.
answered May 4 at 23:52
ClaytonClayton
20.1k33488
$endgroup$
add a comment |
$begingroup$
For any $xinBbb R$, we have $f(x)=f(frac x2+frac x2)=left(f(x/2)right)^2geq0$, so we know $f(x)geq0$ for all $xinBbb R$. Now suppose $f(x_0)=0$. Then $f(0)=f(x_0+(-x_0))=f(x_0)f(-x_0)=0$, which is a contradiction. Thus $f(x)>0$ for all $x$.
answered May 4 at 23:52
ClaytonClayton
20.1k33488
$endgroup$
add a comment |
$begingroup$
For any $xinBbb R$, we have $f(x)=f(frac x2+frac x2)=left(f(x/2)right)^2geq0$, so we know $f(x)geq0$ for all $xinBbb R$. Now suppose $f(x_0)=0$. Then $f(0)=f(x_0+(-x_0))=f(x_0)f(-x_0)=0$, which is a contradiction. Thus $f(x)>0$ for all $x$.
answered May 4 at 23:52
ClaytonClayton
20.1k33488
$endgroup$
For any $xinBbb R$, we have $f(x)=f(frac x2+frac x2)=left(f(x/2)right)^2geq0$, so we know $f(x)geq0$ for all $xinBbb R$. Now suppose $f(x_0)=0$. Then $f(0)=f(x_0+(-x_0))=f(x_0)f(-x_0)=0$, which is a contradiction. Thus $f(x)>0$ for all $x$.
answered May 4 at 23:52
ClaytonClayton
20.1k33488
edited May 5 at 0:10
answered May 4 at 23:52
ClaytonClayton
20.1k33488
answered May 4 at 23:52
ClaytonClayton
20.1k33488
answered May 4 at 23:52
ClaytonClayton
20.1k33488
20.1k33488
add a comment |
add a comment |
$begingroup$
The first step when thinking about a problem like this would be to ask "What would happen if $f(z) = 0$ for some value?". Well since $f(x + y) = f(x)f(y)$ that would mean you could subtract z from any number to find a $f(x)f(z)$ pair that would have to equal zero. So all the numbers would have to be zero, which is contradicted by the questions requirement that $f(0)=1$.
You can make a similar argument for negative numbers. Any positive number times a negative number is a negative number. So if there was a number that make $f(x)$ negative then you could always find a $f(x)f(y)$ that would also be negative and force f(0) to be negative which we know it is not.
answered May 5 at 0:08
Q the PlatypusQ the Platypus
2,9821136
$endgroup$
$begingroup$
I'm not sure what you mean when you say you can make a similar argument for negative numbers. Do you mean that saying something like f(z) = -1 for some value then you would need an x+y pair that = -1? Sorry, I am not good at understanding proofs.
$endgroup$
– Macey
May 5 at 0:25
add a comment |
$begingroup$
The first step when thinking about a problem like this would be to ask "What would happen if $f(z) = 0$ for some value?". Well since $f(x + y) = f(x)f(y)$ that would mean you could subtract z from any number to find a $f(x)f(z)$ pair that would have to equal zero. So all the numbers would have to be zero, which is contradicted by the questions requirement that $f(0)=1$.
You can make a similar argument for negative numbers. Any positive number times a negative number is a negative number. So if there was a number that make $f(x)$ negative then you could always find a $f(x)f(y)$ that would also be negative and force f(0) to be negative which we know it is not.
answered May 5 at 0:08
Q the PlatypusQ the Platypus
2,9821136
$endgroup$
$begingroup$
I'm not sure what you mean when you say you can make a similar argument for negative numbers. Do you mean that saying something like f(z) = -1 for some value then you would need an x+y pair that = -1? Sorry, I am not good at understanding proofs.
$endgroup$
– Macey
May 5 at 0:25
add a comment |
$begingroup$
The first step when thinking about a problem like this would be to ask "What would happen if $f(z) = 0$ for some value?". Well since $f(x + y) = f(x)f(y)$ that would mean you could subtract z from any number to find a $f(x)f(z)$ pair that would have to equal zero. So all the numbers would have to be zero, which is contradicted by the questions requirement that $f(0)=1$.
You can make a similar argument for negative numbers. Any positive number times a negative number is a negative number. So if there was a number that make $f(x)$ negative then you could always find a $f(x)f(y)$ that would also be negative and force f(0) to be negative which we know it is not.
answered May 5 at 0:08
Q the PlatypusQ the Platypus
2,9821136
$endgroup$
The first step when thinking about a problem like this would be to ask "What would happen if $f(z) = 0$ for some value?". Well since $f(x + y) = f(x)f(y)$ that would mean you could subtract z from any number to find a $f(x)f(z)$ pair that would have to equal zero. So all the numbers would have to be zero, which is contradicted by the questions requirement that $f(0)=1$.
You can make a similar argument for negative numbers. Any positive number times a negative number is a negative number. So if there was a number that make $f(x)$ negative then you could always find a $f(x)f(y)$ that would also be negative and force f(0) to be negative which we know it is not.
answered May 5 at 0:08
Q the PlatypusQ the Platypus
2,9821136
edited May 5 at 0:30
answered May 5 at 0:08
Q the PlatypusQ the Platypus
2,9821136
answered May 5 at 0:08
Q the PlatypusQ the Platypus
2,9821136
answered May 5 at 0:08
Q the PlatypusQ the Platypus
2,9821136
2,9821136
$begingroup$
I'm not sure what you mean when you say you can make a similar argument for negative numbers. Do you mean that saying something like f(z) = -1 for some value then you would need an x+y pair that = -1? Sorry, I am not good at understanding proofs.
$endgroup$
– Macey
May 5 at 0:25
add a comment |
$begingroup$
I'm not sure what you mean when you say you can make a similar argument for negative numbers. Do you mean that saying something like f(z) = -1 for some value then you would need an x+y pair that = -1? Sorry, I am not good at understanding proofs.
$endgroup$
– Macey
May 5 at 0:25
$begingroup$
I'm not sure what you mean when you say you can make a similar argument for negative numbers. Do you mean that saying something like f(z) = -1 for some value then you would need an x+y pair that = -1? Sorry, I am not good at understanding proofs.
$endgroup$
– Macey
May 5 at 0:25
$begingroup$
I'm not sure what you mean when you say you can make a similar argument for negative numbers. Do you mean that saying something like f(z) = -1 for some value then you would need an x+y pair that = -1? Sorry, I am not good at understanding proofs.
$endgroup$
– Macey
May 5 at 0:25
add a comment |
$begingroup$
If you diffentiate the equality wrt. $x$, you get $f'(x+y)=f'(x)f(y)$, so for $x=0$ you get $$f'(y)=f(y)$$
You may know that the solutions of this differential equation are of the form $lambda e^x$. Applying $f'(0)=1$ you get $f(x)=e^x$ as a unique solution, and it is indeed $>0$
answered May 4 at 23:58
elidiotelidiot
1,27814
$endgroup$
$begingroup$
How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
$endgroup$
– Macey
May 5 at 0:03
$begingroup$
$f'=f$ is a differential equation, which I solve to answer the question. Read until the end, I can conclude by saying that the exponential is $>0$
$endgroup$
– elidiot
May 5 at 0:05
add a comment |
$begingroup$
If you diffentiate the equality wrt. $x$, you get $f'(x+y)=f'(x)f(y)$, so for $x=0$ you get $$f'(y)=f(y)$$
You may know that the solutions of this differential equation are of the form $lambda e^x$. Applying $f'(0)=1$ you get $f(x)=e^x$ as a unique solution, and it is indeed $>0$
answered May 4 at 23:58
elidiotelidiot
1,27814
$endgroup$
$begingroup$
How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
$endgroup$
– Macey
May 5 at 0:03
$begingroup$
$f'=f$ is a differential equation, which I solve to answer the question. Read until the end, I can conclude by saying that the exponential is $>0$
$endgroup$
– elidiot
May 5 at 0:05
add a comment |
$begingroup$
If you diffentiate the equality wrt. $x$, you get $f'(x+y)=f'(x)f(y)$, so for $x=0$ you get $$f'(y)=f(y)$$
You may know that the solutions of this differential equation are of the form $lambda e^x$. Applying $f'(0)=1$ you get $f(x)=e^x$ as a unique solution, and it is indeed $>0$
answered May 4 at 23:58
elidiotelidiot
1,27814
$endgroup$
If you diffentiate the equality wrt. $x$, you get $f'(x+y)=f'(x)f(y)$, so for $x=0$ you get $$f'(y)=f(y)$$
You may know that the solutions of this differential equation are of the form $lambda e^x$. Applying $f'(0)=1$ you get $f(x)=e^x$ as a unique solution, and it is indeed $>0$
answered May 4 at 23:58
elidiotelidiot
1,27814
answered May 4 at 23:58
elidiotelidiot
1,27814
answered May 4 at 23:58
elidiotelidiot
1,27814
answered May 4 at 23:58
elidiotelidiot
1,27814
1,27814
$begingroup$
How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
$endgroup$
– Macey
May 5 at 0:03
$begingroup$
$f'=f$ is a differential equation, which I solve to answer the question. Read until the end, I can conclude by saying that the exponential is $>0$
$endgroup$
– elidiot
May 5 at 0:05
add a comment |
$begingroup$
How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
$endgroup$
– Macey
May 5 at 0:03
$begingroup$
$f'=f$ is a differential equation, which I solve to answer the question. Read until the end, I can conclude by saying that the exponential is $>0$
$endgroup$
– elidiot
May 5 at 0:05
$begingroup$
How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
$endgroup$
– Macey
May 5 at 0:03
$begingroup$
How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
$endgroup$
– Macey
May 5 at 0:03
$begingroup$
$f'=f$ is a differential equation, which I solve to answer the question. Read until the end, I can conclude by saying that the exponential is $>0$
$endgroup$
– elidiot
May 5 at 0:05
$begingroup$
$f'=f$ is a differential equation, which I solve to answer the question. Read until the end, I can conclude by saying that the exponential is $>0$
$endgroup$
– elidiot
May 5 at 0:05
add a comment |
$begingroup$
Here is a helpful clue. For any base "b". $b^x+y = b^xb^y$
$endgroup$
– Q the Platypus
May 4 at 23:48
$begingroup$
Incidentally, the assumption that $f$ is differentiable everywhere is redundant. Just from the assumption that $f$ is differentiable at $x=0$ we then get $lim_hto 0 fracf(x+h)-f(x)h = lim_hto 0 fracf(x)f(h)-f(x)h = f(x) lim_hto 0 fracf(h)-1h = f(x) f'(0) = f(x)$ exists for every $x$.
$endgroup$
– Daniel Schepler
May 5 at 5:24