F is a Function Everywhere Differentiable With the Following Properties...

**Lord Darkin** · October 7th 2009, 01:59 PM

Question: Let f be a function that is everywhere differentiable and has the following properties.

1.) f(x+h) = (f(x) + f(h)) / (f(-x) + f(-h)) for all real numbers of x and h.
2.) f(x) > 0 for all real numbers of x
3.) f '(0) = -1

PART A
Find the value of f(0).

PART B
Show that f(-x) = 1/f(x) for all real numbers x.

PART C
Using part (b), show that f(x+h) = f(x)f(h) for all real numbers h and x.

PART D
Use the definition of the derivative to find f '(x) in terms of f(x).

Hi guys, this is my first post here.

I have a calculus problem that I'm a bit unsure of how to start, mainly because my teacher is assinging this problem without teaching us the material first (just her style I guess).

I'll give it a shot, anyway. It's not fair that I get help without doing some work.

----------

Because of information 3), I know that f '(0) = 1. Therefore, f (h) = x - 1. I think info 3 is referring to f(h) since info 2 says f(x) is greater than zero for all real numbers, so that would mean it can't be x - 1 because if x=0, f(x) = -1.

But I'm trying to find f(0) ... not sure how to proceed from here or if it refers to f(x) or f(h).

**Enrique2** · October 7th 2009, 02:04 PM

Try to give concrete values, let say put x=0 and h=0 in 1 and just observe what's going on.

**Lord Darkin** · October 7th 2009, 02:10 PM

Originally Posted by Enrique2

Try to give concrete values, let say put x=0 and h=0 in 1 and just observe what's going on.

Okay.

f(0 + 0) = (f(0) + f(0)) / (f(0) f(0)) (Sorry, I haven't memorized the latex codes yet.)

I would assume that the answer is zero, but since f(x) > 0 I can't do that ...

So I'm thinking that f(x+h) must equal negative one (-1) since if f(0) for h is 0, then that leaves f(x)/f(-x).

**tonio** · October 7th 2009, 02:11 PM

Originally Posted by Lord Darkin

Question: Let f be a function that is everywhere differentiable and has the following properties.

1.) f(x+h) = (f(x) + f(h)) / (f(-x) + f(-h)) for all real numbers of x and h.
2.) f(x) > 0 for all real numbers of x
3.) f '(0) = -1

PART A
Find the value of f(0).

== Use (1) with x = h = 0 (hint: you have to use that f(x) > 0 for all x)

PART B
Show that f(-x) = 1/f(x) for all real numbers x.

== Again use (1) with x and h = 0 and solve the easy resulting eq. for f(x)

PART C
Using part (b), show that f(x+h) = f(x)f(h) for all real numbers h and x.

== Use again (1) together with Part B and solve the resulting equation.

PART D
Use the definition of the derivative to find f(x) in terms of f(x).

== I presume you meant here "...to find f(x)' in terms of f(x)"...and this is the most beautiful and elegant part of the darn exercise!
Indeed, use the definition fo derivative and use all the preceeding parts, and ALSO you'll have to use a little arithmetic of limits...

Tonio

Hi guys, this is my first post here.

I have a calculus problem that I'm a bit unsure of how to start, mainly because my teacher is assinging this problem without teaching us the material first (just her style I guess).

I'll give it a shot, anyway. It's not fair that I get help without doing some work.

----------

Because of information 3), I know that f '(0) = 1. Therefore, f (h) = x - 1. I think info 3 is referring to f(h) since info 2 says f(x) is greater than zero for all real numbers, so that would mean it can't be x - 1 because if x=0, f(x) = -1.

But I'm trying to find f(0) ... not sure how to proceed from here or if it refers to f(x) or f(h).

..............................

**Lord Darkin** · October 7th 2009, 02:12 PM

Edit, stupid me ... going to read the post, sorry.

I didn't realize the help was in the quote.

**Lord Darkin** · October 7th 2009, 02:24 PM

The part d of the question was incorrect in my original post - I have now edited it.

Part A Attempt

Let x=h=0

If x=1

f(1 + 0) = (f(1) + f(0)) / (f(-1) + f(0))

f(1) = f(1) / f(-1)

I am stuck here .. I'm not even sure if the work 2 lines above this is correct.

**skeeter** · October 7th 2009, 02:33 PM

Originally Posted by Lord Darkin

Question: Let f be a function that is everywhere differentiable and has the following properties.

1.) f(x+h) = (f(x) + f(h)) / (f(-x) + f(-h)) for all real numbers of x and h.
2.) f(x) > 0 for all real numbers of x
3.) f '(0) = -1

PART A
Find the value of f(0).

PART B
Show that f(-x) = 1/f(x) for all real numbers x.

PART C
Using part (b), show that f(x+h) = f(x)f(h) for all real numbers h and x.

PART D
Use the definition of the derivative to find f(x) in terms of f(x).

Hi guys, this is my first post here.

I have a calculus problem that I'm a bit unsure of how to start, mainly because my teacher is assinging this problem without teaching us the material first (just her style I guess).

I'll give it a shot, anyway. It's not fair that I get help without doing some work.

----------

Because of information 3), I know that f '(0) = 1. Therefore, f (h) = x - 1. I think info 3 is referring to f(h) since info 2 says f(x) is greater than zero for all real numbers, so that would mean it can't be x - 1 because if x=0, f(x) = -1.

But I'm trying to find f(0) ... not sure how to proceed from here or if it refers to f(x) or f(h).

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

part (a) ... let $x = h = 0$ , as suggested

$f(0+0) = f(0) = \frac{2f(0)}{2f(0)} = 1$

part (b) ... let only $h = 0$

$f(x+0) = \frac{f(x) + f(0)}{f(-x) + f(0)}$

$f(x) = \frac{f(x) + 1}{f(-x) + 1}$

$f(-x) + 1 = \frac{f(x)+1}{f(x)}$

$f(-x) + 1 = 1 + \frac{1}{f(x)}$

$f(-x) = \frac{1}{f(x)}$

part (c) ...

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

$f(x+h) = \frac{f(x) + f(h)}{\frac{1}{f(x)} + \frac{1}{f(h)}}$

$f(x+h) = \frac{f(x) + f(h)}{\frac{f(h)+f(x)}{f(x)f(h)}} = f(x)f(h)$

part(d) I'll leave for you to attempt.

**Lord Darkin** · October 7th 2009, 03:07 PM

Originally Posted by skeeter

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

part (a) ... let $x = h = 0$ , as suggested

$f(0+0) = f(0) = \frac{2f(0)}{2f(0)} = 1$

part (b) ... let only $h = 0$

$f(x+0) = \frac{f(x) + f(0)}{f(-x) + f(0)}$

$f(x) = \frac{f(x) + 1}{f(-x) + 1}$

$f(-x) + 1 = \frac{f(x)+1}{f(x)}$

$f(-x) + 1 = 1 + \frac{1}{f(x)}$

$f(-x) = \frac{1}{f(x)}$

part (c) ...

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

$f(x+h) = \frac{f(x) + f(h)}{\frac{1}{f(x)} + \frac{1}{f(h)}}$

$f(x+h) = \frac{f(x) + f(h)}{\frac{f(h)+f(x)}{f(x)f(h)}} = f(x)f(h)$

part(d) I'll leave for you to attempt.

Many thanks, skeeter. I definitely need more practice with these kind of problems but I hope that once I keep doing them, they'll become second nature.

Anyway, I guess it's time to use the definition of the derivative:

$f'(x) = \frac{f(x + h) - f(x)}{h}$

$f'(x) = \frac{f(x)f(h) - f(x)}{h}$

Now, I'm not sure what to do next. tonio said I would have to use limits. However, do I use h -> 0 or h -> infinity? When I use h -> infinity, I end up getting
$f '(x) = f(x)$

which I don't think is right.

**tonio** · October 7th 2009, 03:42 PM

Originally Posted by Lord Darkin

Many thanks, skeeter. I definitely need more practice with these kind of problems but I hope that once I keep doing them, they'll become second nature.

Anyway, I guess it's time to use the definition of the derivative:

$f'(x) = \frac{f(x + h) - f(x)}{h}$

$f'(x) = \frac{f(x)f(h) - f(x)}{h}$

Now, I'm not sure what to do next. tonio said I would have to use limits. However, do I use h -> 0 or h -> infinity? When I use h -> infinity, I end up getting
$f '(x) = f(x)$

which I don't think is right.

You're almost there:

f'(x) = lim f(x)*[(f(h) - 1)/h]....now, what is lim (f(h) - 1}/h taking into consideration the first parts of the question??
All the time above limits are taken with h --> 0

Tonio

**Lord Darkin** · October 7th 2009, 04:04 PM

Originally Posted by tonio

You're almost there:

f'(x) = lim f(x)*[(f(h) - 1)/h]....now, what is lim (f(h) - 1}/h taking into consideration the first parts of the question??
All the time above limits are taken with h --> 0

Tonio

I tried thinking about this for 15 minutes but am stumped.

I had a problem exactly like this before, except the value of lim (f(h)-1)/h was given in the problem.

I'm not sure how to apply the earlier parts of the question.

Am I supposed to do something like this:

$ f'(x) = \frac{f(x + h) - f(x)}{h} $

Let x=1

$ f'(x) = \frac{f(1 + h) - 1}{h} $

$ f'(x) = \frac{1 + f(h) - 1}{h} $

f'(0) = -1

f'(x) = 1 + -2

Is lim ((f(h) -1) / h)negative 2?

**skeeter** · October 7th 2009, 04:05 PM

you were given $f'(0) = -1$

$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$

$-1 = \textcolor{red}{\lim_{x \to 0} \frac{f(x) - 1}{x}}$

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

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

$f'(x) = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h}$

$f'(x) = f(x) \cdot \textcolor{red}{\lim_{h \to 0} \frac{f(h) - 1}{h}}$

$f'(x) = f(x) \cdot \textcolor{red}{(-1)} = -f(x)$

**Lord Darkin** · October 7th 2009, 04:16 PM

Originally Posted by skeeter

you were given $f'(0) = -1$

$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$

$-1 = \textcolor{red}{\lim_{x \to 0} \frac{f(x) - 1}{x}}$

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

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

$f'(x) = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h}$

$f'(x) = f(x) \cdot \textcolor{red}{\lim_{h \to 0} \frac{f(h) - 1}{h}}$

$f'(x) = f(x) \cdot \textcolor{red}{(-1)} = -f(x)$

So my effort was wrong again.

So now I understand what tonio meant ... I just had to go back to the definition of the derivative and look at the third piece ofinfo (f'(0) = -1).

Thanks skeeter, no wonder you're a MHF expert (and probably tonio as well). Thanks to both of you!

I just have a question, though.

When you go from this step to the next, how come the f(x) part goes outside the "limit"?

$f'(x) = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h}$

$f'(x) = f(x) \cdot \textcolor{red}{\lim_{h \to 0} \frac{f(h) - 1}{h}}$

**skeeter** · October 7th 2009, 04:27 PM

Originally Posted by Lord Darkin

I just have a question, though.

When you go from this step to the next, how come the f(x) part goes outside the "limit"?
$f'(x) = \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h}$

$f'(x) = f(x) \cdot \textcolor{red}{\lim_{h \to 0} \frac{f(h) - 1}{h}}$

because the limit is as h approaches 0 , which has no effect on a function of x only, f(x).

**Lord Darkin** · October 7th 2009, 04:32 PM

Originally Posted by skeeter

because the limit is as h approaches 0 , which has no effect on a function of x only, f(x).

Thanks.

Problem completely solved and understood.

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