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Simple Derivative Proof?

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2

2

$begingroup$

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 1 hour ago

Macey

asked 1 hour ago

MaceyMacey

112

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here is a helpful clue. For any base "b". $b^x+y = b^xb^y$
  $endgroup$
  – Q the Platypus
  1 hour ago
  

  
  
  
  

add a comment | 

2

2

$begingroup$

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 1 hour ago

Macey

asked 1 hour ago

MaceyMacey

112

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here is a helpful clue. For any base "b". $b^x+y = b^xb^y$
  $endgroup$
  – Q the Platypus
  1 hour ago
  

  
  
  
  

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 1 hour ago

Macey

asked 1 hour ago

MaceyMacey

112

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 1 hour ago

Macey

asked 1 hour ago

MaceyMacey

112

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 1 hour ago

Macey

asked 1 hour ago

MaceyMacey

112

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 1 hour ago

MaceyMacey

112

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 1 hour ago

MaceyMacey

112

3 Answers
3

active

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4

$begingroup$

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 1 hour ago

answered 1 hour ago

Q the PlatypusQ the Platypus

2,9811135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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 1 hour ago

answered 1 hour ago

ClaytonClayton

20k33388

$endgroup$

add a comment | 

0

$begingroup$

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 1 hour ago

elidiotelidiot

1,21014

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
  $endgroup$
  – Macey
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f'=f$ 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
  1 hour ago
  
  

  
  
  
  

add a comment | 

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4

$begingroup$

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 1 hour ago

answered 1 hour ago

Q the PlatypusQ the Platypus

2,9811135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  
  

  
  
  
  

add a comment | 

4

$begingroup$

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 1 hour ago

answered 1 hour ago

Q the PlatypusQ the Platypus

2,9811135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  
  

  
  
  
  

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 1 hour ago

answered 1 hour ago

Q the PlatypusQ the Platypus

2,9811135

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Q the PlatypusQ the Platypus

2,9811135

answered 1 hour ago

Q the PlatypusQ the Platypus

2,9811135

answered 1 hour ago

Q the PlatypusQ the Platypus

2,9811135

2

$begingroup$

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 1 hour ago

answered 1 hour ago

ClaytonClayton

20k33388

$endgroup$

add a comment | 

2

$begingroup$

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 1 hour ago

answered 1 hour ago

ClaytonClayton

20k33388

$endgroup$

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 1 hour ago

answered 1 hour ago

ClaytonClayton

20k33388

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

ClaytonClayton

20k33388

answered 1 hour ago

ClaytonClayton

20k33388

answered 1 hour ago

ClaytonClayton

20k33388

0

$begingroup$

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 1 hour ago

elidiotelidiot

1,21014

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
  $endgroup$
  – Macey
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f'=f$ 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
  1 hour ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

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 1 hour ago

elidiotelidiot

1,21014

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you know that f(x) must be greater than 0 if f'(y)=f(y)? Thank you.
  $endgroup$
  – Macey
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f'=f$ 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
  1 hour ago
  
  

  
  
  
  

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 1 hour ago

elidiotelidiot

1,21014

$endgroup$

share|cite|improve this answer

answered 1 hour ago

elidiotelidiot

1,21014

answered 1 hour ago

elidiotelidiot

1,21014

answered 1 hour ago

elidiotelidiot

1,21014