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0

3

$begingroup$

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

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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.

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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

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

0

3

$begingroup$

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

share|cite|improve this question

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

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

share|cite|improve this question

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.

$endgroup$

share|cite|improve this question

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.

share|cite|improve this question

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

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

3 Answers
3

active

oldest

votes

9

$begingroup$

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 $x$.

share|cite|improve this answer

edited May 5 at 0:10

answered May 4 at 23:52

ClaytonClayton

20.1k33488

$endgroup$

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

edited May 5 at 0:30

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 | 

1

$begingroup$

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 $>0$

share|cite|improve this answer

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 | 

9

$begingroup$

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 $x$.

share|cite|improve this answer

edited May 5 at 0:10

answered May 4 at 23:52

ClaytonClayton

20.1k33488

$endgroup$

add a comment | 

9

$begingroup$

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 $x$.

share|cite|improve this answer

edited May 5 at 0:10

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 $x$.

share|cite|improve this answer

edited May 5 at 0:10

answered May 4 at 23:52

ClaytonClayton

20.1k33488

$endgroup$

share|cite|improve this answer

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

5

$begingroup$

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.

share|cite|improve this answer

edited May 5 at 0:30

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 | 

5

$begingroup$

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.

share|cite|improve this answer

edited May 5 at 0:30

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.

share|cite|improve this answer

edited May 5 at 0:30

answered May 5 at 0:08

Q the PlatypusQ the Platypus

2,9821136

$endgroup$

share|cite|improve this answer

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

1

$begingroup$

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 $>0$

share|cite|improve this answer

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 | 

1

$begingroup$

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 $>0$

share|cite|improve this answer

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 $>0$

share|cite|improve this answer

answered May 4 at 23:58

elidiotelidiot

1,27814

$endgroup$

share|cite|improve this answer

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